Class 10 Mathematics - NCERT Mathematics
NCERT Solutions for Class 10 Maths Chapter 10: Circles
Chapter 10 Circles focuses on the concept of tangent to a circle and the properties associated with it. Students learn two key theorems: (1) A tangent to a circle is perpendicular to the radius through the point of contact, and (2) The lengths of tangents drawn from an external point to a circle are equal. These theorems form the basis for solving numerical and proof-based problems. This chapter is part of the Geometry unit and carries 4 to 6 marks in the CBSE Class 10 board exam 2027.
Syllabus Update 2026-27
As per the CBSE 2026-27 syllabus, Chapter 10 Circles has 2 exercises - Exercise 10.1 (4 questions on tangent basics and identification) and Exercise 10.2 (13 questions on tangent proofs and numerical problems). This chapter belongs to Unit IV: Geometry, which carries a total of 15 marks in the board exam. Questions from this chapter typically appear as 1 MCQ (1 mark) and 1 short/long answer (3-5 marks).
Table of Contents
10.1Exercise 10.1 - Tangent to a Circle
Q1How many tangents can a circle have?
A circle can have infinitely many tangents.
A tangent is a line that touches the circle at exactly one point. Since a circle has infinitely many points on its circumference, and at each point exactly one tangent can be drawn, a circle has infinitely many tangents.
Q2Fill in the blanks: (i) A tangent to a circle intersects it in ______ point(s). (ii) A line intersecting a circle in two points is called a ______. (iii) A circle can have ______ parallel tangents at the most. (iv) The common point of a tangent to a circle and the circle is called ______.
(i) A tangent to a circle intersects it in one point.
(ii) A line intersecting a circle in two points is called a secant.
(iii) A circle can have two parallel tangents at the most.
(These are the tangents at the two ends of a diameter, which are parallel to each other.)
(iv) The common point of a tangent to a circle and the circle is called the point of contact (or point of tangency).
Q3A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is: (A) 12 cm (B) 13 cm (C) 8.5 cm (D) sqrt(119) cm
Since PQ is a tangent at point P, and OP is the radius:
OP is perpendicular to PQ (tangent is perpendicular to radius at point of contact).
In right triangle OPQ:
OQ^2 = OP^2 + PQ^2 (Pythagoras theorem)
12^2 = 5^2 + PQ^2
144 = 25 + PQ^2
PQ^2 = 119
PQ = sqrt(119)
Answer: (D) sqrt(119) cm
Q4Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Construction steps:
- Draw a circle with centre O and any radius.
- Draw a line 'l' (the given line) at some distance from the circle.
- Draw a line parallel to 'l' that just touches the circle at exactly one point P. This is the tangent.
- Draw another line parallel to 'l' that intersects the circle at two points A and B. This is the secant.
The tangent touches the circle at one point, and the secant cuts through it at two points. Both lines are parallel to the given line 'l'.
Note: This is a construction-based question. Students should draw the figure in their answer sheet showing the circle, the given line, the tangent, and the secant clearly labelled.
10.2Exercise 10.2 - Number of Tangents from a Point to a Circle
Q1From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is: (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm
Let O be the centre and T be the point of contact.
OT is perpendicular to QT (tangent perpendicular to radius).
In right triangle OTQ:
OQ^2 = OT^2 + QT^2
25^2 = OT^2 + 24^2
625 = OT^2 + 576
OT^2 = 49
OT = 7
Answer: (A) 7 cm
Q2In the figure, if TP and TQ are the two tangents to a circle with centre O so that angle POQ = 110 degrees, then angle PTQ is equal to: (A) 60 degrees (B) 70 degrees (C) 80 degrees (D) 90 degrees
Since TP and TQ are tangents from external point T:
- OP is perpendicular to TP, so angle OPT = 90 degrees
- OQ is perpendicular to TQ, so angle OQT = 90 degrees
In quadrilateral OPTQ, the sum of all angles = 360 degrees:
angle OPT + angle PTQ + angle OQT + angle POQ = 360
90 + angle PTQ + 90 + 110 = 360
angle PTQ = 360 - 290
angle PTQ = 70 degrees
Answer: (B) 70 degrees
Q3If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80 degrees, then angle POA is equal to: (A) 50 degrees (B) 60 degrees (C) 70 degrees (D) 80 degrees
Given: angle APB = 80 degrees.
Since PA and PB are tangents from external point P:
- OA is perpendicular to PA (angle OAP = 90 degrees)
- OB is perpendicular to PB (angle OBP = 90 degrees)
In quadrilateral OAPB:
angle OAP + angle APB + angle OBP + angle AOB = 360
90 + 80 + 90 + angle AOB = 360
angle AOB = 100 degrees
Since OP bisects angle AOB (tangents from external point, OP is the line of symmetry):
angle POA = angle AOB / 2 = 100 / 2 = 50 degrees
Answer: (A) 50 degrees
Q4Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Given: AB is a diameter of a circle with centre O. PQ is a tangent at A and RS is a tangent at B.
To prove: PQ is parallel to RS.
Proof:
Since PQ is a tangent at A and OA is the radius:
OA is perpendicular to PQ
So angle OAP = 90 degrees ... (i)
Since RS is a tangent at B and OB is the radius:
OB is perpendicular to RS
So angle OBS = 90 degrees ... (ii)
From (i) and (ii):
angle OAP = angle OBS = 90 degrees
But these are alternate interior angles formed by the transversal AB cutting lines PQ and RS.
Since alternate interior angles are equal, PQ is parallel to RS. ∎
Q5Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Given: A tangent PQ to a circle at point T.
To prove: The perpendicular to PQ at T passes through the centre O.
Proof (by contradiction):
Let us assume the perpendicular at T does not pass through the centre O. Let the perpendicular at T pass through some other point O' (not the centre).
Then angle O'TP = 90 degrees ... (i)
But we know that the tangent at any point is perpendicular to the radius through the point of contact.
So OT is perpendicular to PQ.
Therefore angle OTP = 90 degrees ... (ii)
From (i) and (ii):
angle O'TP = angle OTP = 90 degrees
This is only possible if O' and O coincide (since there can be only one perpendicular to a line at a given point).
This contradicts our assumption that O' is different from O.
Therefore, the perpendicular at the point of contact to the tangent passes through the centre. ∎
Q6The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Let O be the centre and T be the point of contact.
Given: OA = 5 cm (distance from centre), AT = 4 cm (tangent length).
Since OT is perpendicular to AT (tangent perpendicular to radius):
In right triangle OTA:
OA^2 = OT^2 + AT^2
5^2 = OT^2 + 4^2
25 = OT^2 + 16
OT^2 = 9
OT = 3
The radius of the circle is 3 cm.
Q7Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Let O be the common centre. Let AB be a chord of the larger circle that is a tangent to the smaller circle at point P.
Since AB is a tangent to the smaller circle at P:
OP is perpendicular to AB, and OP = 3 cm (radius of smaller circle).
In right triangle OPA:
OA^2 = OP^2 + PA^2
5^2 = 3^2 + PA^2
25 = 9 + PA^2
PA^2 = 16
PA = 4 cm
Since the perpendicular from the centre to a chord bisects the chord:
AB = 2 x PA = 2 x 4 = 8 cm
The length of the chord is 8 cm.
Q8A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.
Given: A quadrilateral ABCD circumscribes a circle with centre O. Let the circle touch AB at P, BC at Q, CD at R, and DA at S.
To prove: AB + CD = AD + BC.
Proof:
Using the property that tangent lengths from an external point to a circle are equal:
From vertex A: AP = AS ... (i)
From vertex B: BP = BQ ... (ii)
From vertex C: CQ = CR ... (iii)
From vertex D: DR = DS ... (iv)
Adding equations (i), (ii), (iii), and (iv):
AP + BP + CQ + DR = AS + BQ + CR + DS
Rearranging:
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC
Hence proved: AB + CD = AD + BC. ∎
Q9In the figure, XY and X'Y' are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X'Y' at B. Prove that angle AOB = 90 degrees.
Given: XY and X'Y' are parallel tangents to a circle with centre O, touching the circle at P and Q respectively. AB is a tangent at C, meeting XY at A and X'Y' at B.
To prove: angle AOB = 90 degrees.
Proof:
Join OP, OC, and OQ.
In triangle OPA and triangle OCA:
- OP = OC (radii)
- AP = AC (tangent lengths from external point A)
- OA = OA (common side)
By SSS congruence: triangle OPA is congruent to triangle OCA.
Therefore: angle POA = angle COA
Let angle POA = angle COA = alpha.
Similarly, in triangle OQB and triangle OCB:
- OQ = OC (radii)
- BQ = BC (tangent lengths from external point B)
- OB = OB (common side)
By SSS congruence: triangle OQB is congruent to triangle OCB.
Therefore: angle QOB = angle COB
Let angle QOB = angle COB = beta.
Since PQ is a diameter (XY and X'Y' are parallel tangents, so P and Q are diametrically opposite):
angle POQ = 180 degrees (straight line)
angle POA + angle AOC + angle COB + angle BOQ = 180
alpha + alpha + beta + beta = 180
2(alpha + beta) = 180
alpha + beta = 90
Now, angle AOB = angle AOC + angle COB = alpha + beta = 90 degrees.
Hence proved: angle AOB = 90 degrees. ∎
Q10Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Given: PA and PB are tangents from an external point P to a circle with centre O. A and B are points of contact.
To prove: angle APB + angle AOB = 180 degrees.
Proof:
Since PA is a tangent at A:
OA is perpendicular to PA
angle OAP = 90 degrees ... (i)
Since PB is a tangent at B:
OB is perpendicular to PB
angle OBP = 90 degrees ... (ii)
In quadrilateral OAPB, the sum of all angles = 360 degrees:
angle OAP + angle APB + angle OBP + angle AOB = 360
Substituting from (i) and (ii):
90 + angle APB + 90 + angle AOB = 360
angle APB + angle AOB = 360 - 180
angle APB + angle AOB = 180 degrees
Hence, angle APB and angle AOB are supplementary. ∎
Q11Prove that the parallelogram circumscribing a circle is a rhombus.
Given: ABCD is a parallelogram circumscribing a circle.
To prove: ABCD is a rhombus (all sides are equal).
Proof:
Let the circle touch AB at P, BC at Q, CD at R, and DA at S.
Since tangent lengths from an external point are equal:
AP = AS ... (i)
BP = BQ ... (ii)
CR = CQ ... (iii)
DR = DS ... (iv)
Adding all four equations:
AP + BP + CR + DR = AS + BQ + CQ + DS
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC ... (v)
Since ABCD is a parallelogram:
AB = CD and AD = BC ... (vi)
Substituting (vi) in (v):
AB + AB = AD + AD
2AB = 2AD
AB = AD
Since AB = CD and AD = BC (opposite sides of parallelogram) and AB = AD:
AB = BC = CD = DA
Therefore, ABCD is a rhombus. ∎
Q12A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.
Let the incircle touch BC at D, CA at E, and AB at F.
Given: BD = 8 cm, DC = 6 cm, radius r = 4 cm.
Using equal tangent lengths from external points:
- From B: BF = BD = 8 cm
- From C: CE = CD = 6 cm
- From A: AF = AE = x cm (let)
So: AB = AF + BF = x + 8
AC = AE + CE = x + 6
BC = BD + DC = 8 + 6 = 14 cm
Semi-perimeter: s = (AB + BC + CA)/2 = (x + 8 + 14 + x + 6)/2 = (2x + 28)/2 = x + 14
Area of triangle using incircle formula:
Area = r x s = 4(x + 14) = 4x + 56 ... (I)
Area using Heron's formula:
s = x + 14, s - a = x + 14 - 14 = x, s - b = x + 14 - (x + 6) = 8, s - c = x + 14 - (x + 8) = 6
Area = sqrt[s(s-a)(s-b)(s-c)] = sqrt[(x+14)(x)(8)(6)] = sqrt[48x(x+14)] ... (II)
Equating (I) and (II):
4x + 56 = sqrt[48x(x + 14)]
Squaring both sides:
(4x + 56)^2 = 48x(x + 14)
16(x + 14)^2 = 48x(x + 14)
Divide both sides by (x + 14) [since x + 14 > 0]:
16(x + 14) = 48x
16x + 224 = 48x
224 = 32x
x = 7
Therefore:
AB = x + 8 = 7 + 8 = 15 cm
AC = x + 6 = 7 + 6 = 13 cm
Q13Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Given: ABCD is a quadrilateral circumscribing a circle with centre O. The circle touches AB at P, BC at Q, CD at R, and DA at S.
To prove: angle AOB + angle COD = 180 degrees and angle AOD + angle BOC = 180 degrees.
Proof:
Join OP, OQ, OR, OS, OA, OB, OC, OD.
In triangle OAP and triangle OAS:
- OP = OS (radii)
- AP = AS (tangent lengths from A)
- OA = OA (common)
By SSS: triangle OAP is congruent to triangle OAS
So angle AOP = angle AOS = angle 1 (say)
Similarly:
In triangles OBP and OBQ: angle BOP = angle BOQ = angle 2
In triangles OCQ and OCR: angle COQ = angle COR = angle 3
In triangles ODR and ODS: angle DOR = angle DOS = angle 4
The angles around O sum to 360 degrees:
2(angle 1) + 2(angle 2) + 2(angle 3) + 2(angle 4) = 360
angle 1 + angle 2 + angle 3 + angle 4 = 180 ... (*)
Now:
angle AOB = angle 1 + angle 2
angle COD = angle 3 + angle 4
From (*):
angle AOB + angle COD = (angle 1 + angle 2) + (angle 3 + angle 4) = 180 degrees
Similarly:
angle AOD = angle 1 + angle 4
angle BOC = angle 2 + angle 3
angle AOD + angle BOC = (angle 1 + angle 4) + (angle 2 + angle 3) = 180 degrees
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre. ∎
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Ask AI TutorPrevious Year Board Exam Questions
Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that angle PTQ = 2 angle OPQ.
Since TP and TQ are tangents from T, we have TP = TQ (equal tangent lengths). So triangle TPQ is isosceles. Let angle TPQ = angle TQP = alpha. Then angle PTQ = 180 - 2alpha. Since OP is perpendicular to TP, angle OPT = 90 degrees. So angle OPQ = 90 - alpha. Now, angle PTQ = 180 - 2alpha = 2(90 - alpha) = 2 x angle OPQ. Hence angle PTQ = 2 angle OPQ is proved.
Prove that the lengths of tangents drawn from an external point to a circle are equal.
Given PA and PB are tangents from external point P to a circle with centre O, touching at A and B. Join OA, OB, and OP. Since tangent is perpendicular to radius, angle OAP = angle OBP = 90 degrees. In right triangles OAP and OBP: OA = OB (radii), OP = OP (common hypotenuse), and angle OAP = angle OBP = 90 degrees. By RHS congruence, triangle OAP is congruent to triangle OBP. Therefore PA = PB by CPCT. Hence the tangent lengths from an external point are equal.
In the given figure, a circle is inscribed in triangle ABC with AB = 12 cm, BC = 8 cm, and AC = 10 cm. Find the lengths of AD, BE, and CF where D, E, F are points of contact on AB, BC, and CA respectively.
Let AD = AF = x, BD = BE = y, CE = CF = z (tangent lengths from the same external point are equal). From the sides: x + y = 12 (AB), y + z = 8 (BC), x + z = 10 (AC). Adding all three equations: 2(x + y + z) = 30, so x + y + z = 15. Using x + y = 12 gives z = 3. Using y + z = 8 gives y = 5. Using x + z = 10 gives x = 7. Therefore AD = 7 cm, BE = 5 cm, CF = 3 cm.
Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Given a circle with centre O and a tangent XY at point P on the circle. Take any point Q on XY other than P. Since the tangent touches the circle only at P, the point Q must lie outside the circle. Therefore OQ is greater than OP, where OP is the radius. This is true for every point Q on the line XY other than P. Thus OP is the shortest distance from the centre O to the line XY. Since the shortest distance from a point to a line is the perpendicular distance, OP must be perpendicular to XY. Hence the tangent at any point is perpendicular to the radius through the point of contact.
From an external point P, two tangents PA and PB are drawn to a circle with centre O. If angle APB = 60 degrees, find angle AOB.
Since PA and PB are tangents from the external point P, angle OAP = angle OBP = 90 degrees (tangent is perpendicular to the radius at the point of contact). In quadrilateral OAPB, the sum of all angles equals 360 degrees. So angle OAP + angle APB + angle OBP + angle AOB = 360. Substituting the known values: 90 + 60 + 90 + angle AOB = 360. Therefore angle AOB = 360 - 240 = 120 degrees.
Prove that the parallelogram circumscribing a circle is a rhombus.
Let ABCD be a parallelogram circumscribing a circle. Let the circle touch AB at P, BC at Q, CD at R, and DA at S. By equal tangent lengths from each vertex: AP = AS, BP = BQ, CR = CQ, and DR = DS. Adding all four: (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ), which gives AB + CD = AD + BC. Since ABCD is a parallelogram, AB = CD and AD = BC. So 2AB = 2AD, meaning AB = AD. Since all sides are equal (AB = BC = CD = DA), ABCD is a rhombus. Hence proved.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Let O be the common centre and AB be a chord of the larger circle that is tangent to the smaller circle at P. Since AB is tangent to the smaller circle, OP is perpendicular to AB and OP equals 3 cm. In right triangle OPA, OA = 5 cm (radius of larger circle). By Pythagoras theorem: PA squared = OA squared minus OP squared = 25 minus 9 = 16, so PA = 4 cm. Since the perpendicular from the centre bisects the chord, AB = 2 times PA = 8 cm.
In a right triangle ABC, right-angled at B, BC = 7 cm, AB = 24 cm. A circle is inscribed in the triangle. Find the radius of the circle.
First find AC using Pythagoras theorem: AC = sqrt(AB^2 + BC^2) = sqrt(576 + 49) = sqrt(625) = 25 cm. Let r be the inradius. The incircle touches AB at F, BC at D, and AC at E. Using tangent lengths: BD = BF = r (since angle B = 90 degrees and the incircle touches both sides at distance r from B). Also CD = CE = BC minus r = 7 minus r, and AF = AE = AB minus r = 24 minus r. Now AC = AE + CE, so 25 = (24 minus r) + (7 minus r) = 31 minus 2r. Therefore 2r = 6, giving r = 3 cm.
If two tangents inclined at an angle of 60 degrees are drawn to a circle of radius 3 cm, find the length of each tangent.
Let PA and PB be tangents from external point P to a circle with centre O and radius 3 cm. Given angle APB = 60 degrees. Since OP bisects angle APB, angle OPA = 30 degrees. In right triangle OAP where angle OAP = 90 degrees: tan(30) = OA/PA, which gives 1/sqrt(3) = 3/PA. Solving, PA = 3 times sqrt(3). Therefore the length of each tangent is 3sqrt(3) cm, which is approximately 5.196 cm.
If a circle touches all four sides of a quadrilateral ABCD, prove that AB + CD = BC + DA.
Let the circle touch AB at P, BC at Q, CD at R, and DA at S. Using the theorem that tangent lengths from an external point are equal: AP = AS (from vertex A), BP = BQ (from vertex B), CQ = CR (from vertex C), and DR = DS (from vertex D). Adding all four equalities: AP + BP + CQ + DR = AS + BQ + CR + DS. Rearranging the left and right sides: (AP + BP) + (CQ + DR) = (AS + DS) + (BQ + CQ). This simplifies to AB + CD = DA + BC. Hence proved.
Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.
Let M be the mid-point of arc AB of a circle with centre O. Let PQ be the tangent at M. Join OA, OB, and OM. Since M is the mid-point of arc AB, arc AM = arc BM, so the chords AM = BM (equal arcs subtend equal chords). Also angle AOM = angle BOM, meaning OM bisects angle AOB. Since the perpendicular from the centre to a chord bisects the chord, and OM is the axis of symmetry for chord AB, the line OM is perpendicular to AB. Since OM is also perpendicular to tangent PQ at M, both AB and PQ are perpendicular to OM. Therefore PQ is parallel to AB.
A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.
Let the incircle touch AB at P, BC at Q, CD at R, and DA at S. By the property of equal tangent lengths from an external point: AP = AS (tangents from A), BP = BQ (tangents from B), CQ = CR (tangents from C), and DR = DS (tangents from D). Adding all four: (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ). This directly simplifies to AB + CD = AD + BC, which is the required result. Hence proved.
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Practice PYQ PatternsAssertion-Reason Questions
(A) Both true, Reason explains Assertion · (B) Both true, Reason doesn't explain · (C) Assertion true, Reason false · (D) Assertion false, Reason true
AR1
Assertion (A): A tangent to a circle is perpendicular to the radius through the point of contact.
Reason (R): The shortest distance from the centre of a circle to the tangent is the radius.
Answer: (A)
Both the assertion and reason are true. The reason correctly explains the assertion. Since the radius is the shortest distance from the centre to the tangent line, and the shortest distance from a point to a line is the perpendicular distance, the tangent must be perpendicular to the radius at the point of contact.
AR2
Assertion (A): The lengths of tangents drawn from an external point to a circle are equal.
Reason (R): Tangents drawn from an external point are parallel to each other.
Answer: (C)
The assertion is TRUE - tangent lengths from an external point are indeed equal (proved using RHS congruence of the two right triangles formed). The reason is FALSE - tangents from an external point are NOT parallel; they meet at the external point and diverge outward. The correct reason involves congruence of the two triangles formed with the radii.
AR3
Assertion (A): If two tangents are drawn to a circle from an external point, then the angle between them is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
Reason (R): In quadrilateral OAPB formed by two tangents PA and PB and radii OA and OB, the sum of all angles is 360 degrees.
Answer: (A)
Both are true and the reason explains the assertion. In quadrilateral OAPB, angle OAP = angle OBP = 90 degrees (tangent perpendicular to radius). So angle APB + angle AOB = 360 - 90 - 90 = 180 degrees. Hence the two angles are supplementary, as the assertion states.
AR4
Assertion (A): A circle can have at most two parallel tangents.
Reason (R): Parallel tangents can only be drawn at the two end points of a diameter.
Answer: (A)
Both are true and the reason explains the assertion. Two tangents at the ends of a diameter are perpendicular to the diameter and hence parallel to each other. Any third tangent at a different point would not be parallel to both of them, so the maximum number of parallel tangents is two.
AR5
Assertion (A): The parallelogram circumscribing a circle is always a rhombus.
Reason (R): For a quadrilateral circumscribing a circle, the sum of opposite sides are equal.
Answer: (A)
Both are true and the reason explains the assertion. For a circumscribed quadrilateral: AB + CD = AD + BC. In a parallelogram, AB = CD and AD = BC. Substituting: 2AB = 2AD, so AB = AD. Combined with the parallelogram properties, all four sides are equal, making it a rhombus.
AR6
Assertion (A): From a point inside a circle, exactly one tangent can be drawn to the circle.
Reason (R): A tangent to a circle touches the circle at exactly one point.
Answer: (D)
The assertion is FALSE - from a point inside a circle, NO tangent can be drawn at all (not one). Any line through an interior point of a circle intersects the circle in two points, making it a secant. The reason is TRUE as a correct definition of a tangent. Since the assertion is false and the reason is true, the answer is (D).
Case Study Questions
A landscape architect is designing a park with a circular fountain at its centre. The fountain has a radius of 7 m. A straight pathway touches the boundary of the fountain at point P. Two benches A and B are placed on this pathway on either side of point P at equal distances. The distance from each bench to the centre of the fountain is 25 m.
(i) What is the distance from each bench to the point P where the pathway touches the fountain? (1 mark)
Let O be the centre of the fountain. Since the pathway is a tangent at P, OP is perpendicular to the pathway and OP = 7 m.
For bench A, OA = 25 m. In right triangle OPA:
AP^2 = OA^2 - OP^2 = 625 - 49 = 576
AP = 24 m.
Each bench is 24 m from point P.
(ii) What is the total distance between the two benches A and B? (1 mark)
Since both benches are placed at equal distances from P on either side of the tangent point:
AP = PB = 24 m.
Total distance AB = AP + PB = 24 + 24 = 48 m.
(iii)(a) If another straight pathway from bench A touches the fountain at point Q, what is the length AQ? (2 marks)
Since AP and AQ are both tangents drawn from the same external point A to the circular fountain:
AQ = AP (tangent lengths from an external point are equal)
AQ = 24 m.
(iii)(b) Find the angle subtended by the line segment PQ at the centre O, if angle PAQ = 60 degrees. (2 marks)
Since AP and AQ are tangents from external point A:
angle OPA = angle OQA = 90 degrees (tangent perpendicular to radius).
In quadrilateral OPAQ:
angle OPA + angle PAQ + angle OQA + angle POQ = 360
90 + 60 + 90 + angle POQ = 360
angle POQ = 360 - 240 = 120 degrees.
In a factory, a conveyor belt system uses a large circular pulley of radius 5 cm. An engineer needs to perform various calculations for maintenance purposes. The centre of the pulley is at point O. A quality inspector stands at an external point P from where two belts (tangent lines) PA and PB touch the pulley at points A and B respectively.
(i) If the distance OP from the inspector to the centre of the pulley is 13 cm, find the length of each tangent belt from P to the pulley. (1 mark)
In right triangle OAP (angle OAP = 90 degrees, tangent perpendicular to radius):
PA^2 = OP^2 - OA^2 = 13^2 - 5^2 = 169 - 25 = 144
PA = 12 cm.
Each tangent belt from P to the pulley is 12 cm long.
(ii) Find the angle AOB if the angle between the two tangent belts (angle APB) is 60 degrees. (1 mark)
Since angle APB + angle AOB = 180 degrees (angle between tangents and angle at centre are supplementary):
angle AOB = 180 - 60 = 120 degrees.
(iii)(a) A metal frame shaped as a quadrilateral ABCD circumscribes the pulley, with AB = 12 cm, BC = 9 cm, and CD = 7 cm. Find DA. (2 marks)
For a quadrilateral circumscribing a circle:
AB + CD = BC + DA
12 + 7 = 9 + DA
19 = 9 + DA
DA = 10 cm.
DA = 10 cm.
(iii)(b) If two tangent belts from the inspector at P make an angle of 90 degrees between them, find the distance OP from the inspector to the centre of the pulley. (2 marks)
Given: angle APB = 90 degrees, radius OA = 5 cm.
Since OP bisects angle APB, angle OPA = 45 degrees.
In right triangle OAP (angle OAP = 90 degrees):
sin(45) = OA/OP
1/sqrt(2) = 5/OP
OP = 5sqrt(2) cm.
OP = 5sqrt(2) cm, which is approximately 7.07 cm.
MCQ Practice
MCQ1. If the angle between two tangents drawn from a point P to a circle of radius r and centre O is 90 degrees, then OP is equal to:
(A) r sqrt(2)
(B) 2r
(C) r/sqrt(2)
(D) r
Answer: (A) - Let tangents PA and PB make angle APB = 90 degrees. Since angle OAP = 90 degrees and OP bisects angle APB, angle OPA = 45 degrees. In right triangle OAP: sin(45) = OA/OP, so 1/sqrt(2) = r/OP, giving OP = r sqrt(2).
MCQ2. The length of a tangent from a point A at a distance of 10 cm from the centre of a circle is 8 cm. The radius of the circle is:
(A) 4 cm
(B) 5 cm
(C) 6 cm
(D) 7 cm
Answer: (C) - Using Pythagoras theorem in the right triangle formed by the radius, tangent, and the line joining the centre to the external point: radius^2 = 10^2 - 8^2 = 100 - 64 = 36. So radius = 6 cm.
MCQ3. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is:
(A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
Answer: (A) - Using Pythagoras theorem: r^2 = 25^2 - 24^2 = 625 - 576 = 49. So r = 7 cm.
MCQ4. In a circle of radius 5 cm, AB is a chord of length 8 cm. The tangents at A and B intersect at point P. The length PA is:
(A) 20/3 cm
(B) 25/3 cm
(C) 8/3 cm
(D) 10/3 cm
Answer: (A) - Let M be the midpoint of AB. Then AM = 4 cm and OM = sqrt(25 - 16) = 3 cm. Since OP passes through M and angle OAP = 90 degrees, using OA^2 = OM x OP gives OP = 25/3. Then PA^2 = OP^2 - OA^2 = 625/9 - 25 = 625/9 - 225/9 = 400/9. So PA = 20/3 cm.
MCQ5. If TP and TQ are the two tangents to a circle with centre O so that angle POQ = 110 degrees, then angle PTQ is:
(A) 60 degrees
(B) 70 degrees
(C) 80 degrees
(D) 90 degrees
Answer: (B) - In quadrilateral OPTQ: angle OPT + angle PTQ + angle OQT + angle POQ = 360. So 90 + angle PTQ + 90 + 110 = 360, which gives angle PTQ = 70 degrees.
MCQ6. The number of tangents that can be drawn from a point inside a circle is:
(A) 0
(B) 1
(C) 2
(D) infinite
Answer: (A) - From a point inside a circle, no tangent can be drawn. Any line through an interior point of a circle intersects the circle at two points, making it a secant, not a tangent.
MCQ7. A tangent PQ at point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 13 cm. Length PQ is:
(A) 10 cm
(B) 11 cm
(C) 12 cm
(D) 13 cm
Answer: (C) - PQ is perpendicular to OP (tangent perpendicular to radius). By Pythagoras: PQ^2 = OQ^2 - OP^2 = 169 - 25 = 144. PQ = 12 cm.
MCQ8. If angle between two radii of a circle is 130 degrees, the angle between the tangents at the ends of the radii is:
(A) 90 degrees
(B) 50 degrees
(C) 40 degrees
(D) 70 degrees
Answer: (B) - The angle between tangents and the angle between radii at the points of contact are supplementary (they add up to 180 degrees). So angle between tangents = 180 - 130 = 50 degrees.
MCQ9. If a parallelogram circumscribes a circle, then it is a:
(A) rectangle
(B) square
(C) rhombus
(D) trapezium
Answer: (C) - For a circumscribed quadrilateral: AB + CD = AD + BC. In a parallelogram, AB = CD and AD = BC. So 2AB = 2AD, giving AB = AD. Since all four sides are equal, it is a rhombus.
MCQ10. In the given figure, PA and PB are tangents to the circle with centre O. If angle APB = 50 degrees, then angle OAB is:
(A) 25 degrees
(B) 30 degrees
(C) 35 degrees
(D) 40 degrees
Answer: (A) - angle AOB = 180 - 50 = 130 degrees (supplementary). Since OA = OB (radii), triangle OAB is isosceles. angle OAB = angle OBA = (180 - 130)/2 = 50/2 = 25 degrees.
MCQ11. Two tangents PA and PB are drawn from an external point P. If PA = 4 cm, then PB is:
(A) 2 cm
(B) 4 cm
(C) 8 cm
(D) cannot be determined
Answer: (B) - Tangent lengths from the same external point to a circle are always equal. Since PA = 4 cm, PB must also equal 4 cm.
MCQ12. If a quadrilateral ABCD is circumscribed about a circle with AB = 6 cm, BC = 7 cm, and CD = 4 cm, then DA is:
(A) 3 cm
(B) 4 cm
(C) 5 cm
(D) 7 cm
Answer: (A) - For a circumscribed quadrilateral: AB + CD = BC + DA. So 6 + 4 = 7 + DA. Therefore 10 = 7 + DA, giving DA = 3 cm.
Try Quiz mode - 10 MCQs per topic with instant feedback and scoring.
Start Quiz ModeImportant Questions - Mark-wise
1 Mark Questions
If two tangents are drawn from an external point P to a circle with centre O and radius r such that OP = 2r, find the angle between the two tangents.
In right triangle OAP, sin(angle OPA) = OA/OP = r/2r = 1/2, so angle OPA = 30 degrees. Since OP bisects angle APB, angle APB = 2 x 30 = **60 degrees**.
If a chord of length 8 cm is drawn in a circle of radius 5 cm, what is the distance of the chord from the centre?
Half chord = 4 cm. Distance = sqrt(5^2 - 4^2) = sqrt(25 - 16) = sqrt(9) = **3 cm**.
How many tangents can be drawn to a circle from a point on the circle?
Exactly **one** tangent can be drawn to a circle from a point lying on the circle.
If PA and PB are two tangents from an external point P to a circle with centre O, and angle AOB = 120 degrees, find angle APB.
Since angle APB + angle AOB = 180 degrees (supplementary), angle APB = 180 - 120 = **60 degrees**.
A tangent is drawn from a point at a distance of 17 cm from the centre of a circle of radius 8 cm. Find the length of the tangent.
Tangent length = sqrt(17^2 - 8^2) = sqrt(289 - 64) = sqrt(225) = **15 cm**.
2-3 Mark Questions
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Let AB be a diameter with tangents PQ at A and RS at B. Since PQ is tangent at A, OA perpendicular to PQ gives angle OAP = 90 degrees. Since RS is tangent at B, OB perpendicular to RS gives angle OBS = 90 degrees. Now angle OAP and angle OBS are alternate interior angles formed by transversal AB. Since they are equal (both 90 degrees), PQ is parallel to RS.
Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that angle PTQ = 2 angle OPQ.
Since TP = TQ (equal tangent lengths), triangle TPQ is isosceles with angle TPQ = angle TQP = alpha. So angle PTQ = 180 - 2alpha. Since OP perpendicular to TP, angle OPT = 90 degrees. Thus angle OPQ = angle OPT - angle TPQ = 90 - alpha. Therefore angle PTQ = 180 - 2alpha = 2(90 - alpha) = 2 angle OPQ.
From an external point P, tangents PA and PB are drawn to a circle with centre O. If angle PAB = 25 degrees, find angle APB and angle AOB.
Since PA = PB (equal tangent lengths), triangle PAB is isosceles with angle PAB = angle PBA = 25 degrees. So angle APB = 180 - 25 - 25 = 130 degrees. Since angle APB + angle AOB = 180 degrees (supplementary), angle AOB = 180 - 130 = 50 degrees.
In the figure, a circle touches all four sides of a quadrilateral ABCD. If AB = 6 cm, BC = 9 cm, and CD = 8 cm, find the length of AD.
For a quadrilateral circumscribing a circle, the sum of opposite sides are equal: AB + CD = BC + AD. Substituting: 6 + 8 = 9 + AD. So 14 = 9 + AD, giving AD = 5 cm.
3-4 Mark Questions
Prove that the lengths of tangents drawn from an external point to a circle are equal.
Given: PA and PB are tangents from external point P to a circle with centre O. Join OA, OB, OP.
Since tangent is perpendicular to radius: angle OAP = angle OBP = 90 degrees.
In right triangles OAP and OBP:
- OA = OB (radii of same circle)
- OP = OP (common hypotenuse)
- angle OAP = angle OBP = 90 degrees
By RHS congruence criterion: triangle OAP is congruent to triangle OBP.
By CPCT: PA = PB. Hence proved.
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.
Using equal tangent lengths: BF = BD = 8, CE = CD = 6, AF = AE = x. So AB = x + 8, AC = x + 6, BC = 14. Semi-perimeter s = x + 14. Area by incircle formula = 4(x + 14). Area by Heron's formula = sqrt[(x+14)(x)(8)(6)] = sqrt[48x(x+14)]. Equating and squaring: 16(x+14)^2 = 48x(x+14). Dividing by (x+14): 16(x+14) = 48x, so 16x + 224 = 48x, giving 32x = 224, x = 7. AB = 15 cm, AC = 13 cm.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Let circle touch AB at P, BC at Q, CD at R, DA at S. Join OP, OQ, OR, OS, OA, OB, OC, OD. By SSS congruence of triangles formed at each vertex: angle AOP = angle AOS = a1, angle BOP = angle BOQ = a2, angle COQ = angle COR = a3, angle DOR = angle DOS = a4. Sum of all angles around O: 2(a1 + a2 + a3 + a4) = 360, so a1 + a2 + a3 + a4 = 180. Now angle AOB + angle COD = (a1 + a2) + (a3 + a4) = 180 degrees. Similarly angle AOD + angle BOC = (a1 + a4) + (a2 + a3) = 180 degrees. Hence proved.
Key Concepts
Important Formulas
Study Plan - Circles
Tangent Basics and Theorem 1
Read NCERT Section 10.1 and 10.2. Understand the definition of tangent, secant, and point of contact. Learn Theorem 10.1 (tangent perpendicular to radius) with its proof. Solve Exercise 10.1 (Q1-Q4).
Theorem 2 and Basic Numericals
Study Theorem 10.2 (equal tangent lengths from external point) with its proof. Practice Exercise 10.2 Q1-Q6 focusing on Pythagoras theorem applications in tangent-radius problems.
Proof-Based Questions
Master all proof-based problems from Exercise 10.2 (Q4, Q5, Q8, Q9, Q10, Q11, Q13). Practice writing structured proofs with Given, To Prove, Construction, and Proof steps.
Advanced Problems and PYQs
Solve Exercise 10.2 Q12 (incircle in triangle). Attempt all previous year questions and case study questions. Focus on circumscribed quadrilateral properties and supplementary angle proofs.
Revision and Mock Test
Revise all theorems and their proofs. Solve MCQs and Assertion-Reason questions under timed conditions. Take a chapter-wise mock test covering all question types. Review common mistakes.
Common Mistakes to Avoid
- ✗Forgetting that the tangent is perpendicular to the radius at the point of contact, leading to incorrect angle calculations in quadrilateral OAPB problems
- ✗Confusing the supplementary relationship: angle APB + angle AOB = 180 degrees applies only when PA and PB are tangents from the same external point P
- ✗Incorrectly applying the circumscribed quadrilateral property - the correct relation is AB + CD = AD + BC (opposite sides sum), not adjacent sides
- ✗Not recognizing that tangent lengths from the same external point are equal, which leads to errors in finding unknown side lengths of circumscribed polygons
- ✗Making errors in the incircle problem (Ex 10.2 Q12) by incorrectly setting up Heron's formula or forgetting to use Area = r x s for the incircle area relationship
Tips for Board Exam
- ✓Always draw a clear, labelled diagram showing the circle, tangent, radius, and external point before attempting any problem - this helps visualize the right angles and equal lengths
- ✓Memorize both theorems with proofs: tangent perpendicular to radius (proof by shortest distance) and equal tangent lengths (proof by RHS congruence) as 3-5 mark questions frequently test these
- ✓For angle-finding problems, always start by marking the two 90-degree angles where tangents meet radii, then use the angle sum property of the quadrilateral OAPB (360 degrees)
- ✓In circumscribed polygon problems, label the tangent lengths using variables (e.g., AP = AS = x, BP = BQ = y) right at the start to set up equations systematically
- ✓Practice writing proofs in the structured format: Given, To Prove, Construction (if any), and Proof - CBSE examiners award step marks, so show all intermediate steps clearly
Practice Circles with AI Tutor
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Start Circles - Choose Your ModeAbout Circles - Class 10 Mathematics
Introduction to Circles and Tangents
Chapter 10 of the NCERT Class 10 Mathematics textbook deals with an important topic in geometry: the tangent to a circle and its properties. In earlier classes, students studied circles, their chords, arcs, and angles subtended by them. In Class 10, the focus shifts to tangent lines and the theorems associated with them. A tangent to a circle is a straight line that touches the circle at exactly one point. This point is called the point of contact or the point of tangency. Unlike a secant, which intersects a circle at two points, a tangent merely grazes the surface of the circle.
Tangent and Its Relationship with the Radius
The first major result in this chapter is Theorem 10.1, which states that the tangent at any point of a circle is perpendicular to the radius through the point of contact. This means if you draw a radius from the centre O to the point where a tangent touches the circle at point P, then the angle between OP and the tangent line is exactly 90 degrees. The proof of this theorem relies on the idea that the radius is the shortest distance from the centre to the tangent line. Any other line segment from O to a point on the tangent (other than P) would be longer than the radius because that point lies outside the circle. Since the shortest distance from a point to a line is the perpendicular distance, OP must be perpendicular to the tangent. This theorem is fundamental and is used in almost every problem in the chapter.
Number of Tangents from a Point
The number of tangents that can be drawn to a circle from a given point depends on the position of that point relative to the circle. If the point lies inside the circle, no tangent can be drawn because every line through an interior point intersects the circle at two points (making it a secant). If the point lies on the circle, exactly one tangent can be drawn at that point. If the point lies outside the circle, exactly two tangents can be drawn from that point. These two tangents from an external point have a special property described by the second theorem.
Equal Tangent Lengths from an External Point
Theorem 10.2 states that the lengths of tangents drawn from an external point to a circle are equal. If PA and PB are tangents from an external point P to a circle with centre O, touching the circle at A and B respectively, then PA = PB. The proof uses the RHS (Right angle-Hypotenuse-Side) congruence criterion. In the two right triangles OAP and OBP, the hypotenuse OP is common, OA = OB (both are radii), and both triangles are right-angled at A and B respectively (tangent perpendicular to radius). By RHS congruence, the two triangles are congruent, and therefore PA = PB by CPCT (Corresponding Parts of Congruent Triangles). This theorem also implies that OP bisects angle APB and angle AOB, and that OP is the perpendicular bisector of the chord AB.
The Supplementary Angle Property
An important result that follows from the tangent-radius perpendicularity is that the angle between the two tangents drawn from an external point and the angle subtended by the line segment joining the points of contact at the centre are supplementary. In other words, angle APB + angle AOB = 180 degrees. This can be proved using the angle sum property of quadrilateral OAPB, where angle OAP = angle OBP = 90 degrees. Adding all four angles gives 90 + angle APB + 90 + angle AOB = 360, which simplifies to angle APB + angle AOB = 180 degrees.
Circumscribed Quadrilaterals and Polygons
When a quadrilateral ABCD is drawn to circumscribe a circle (meaning the circle touches all four sides), an important property emerges: the sum of opposite sides are equal, that is, AB + CD = AD + BC. This is proved by assigning tangent length variables at each vertex and using the equal tangent length theorem. For instance, if the circle touches AB at P, BC at Q, CD at R, and DA at S, then AP = AS, BP = BQ, CQ = CR, and DR = DS. Adding these gives AB + CD = AD + BC. A special case of this result is that a parallelogram circumscribing a circle must be a rhombus, because the additional constraint of equal opposite sides forces all four sides to be equal.
Applications and Problem-Solving Techniques
The chapter includes several types of problems that appear regularly in board examinations. Numerical problems typically involve finding the length of a tangent using the Pythagoras theorem in the right triangle formed by the radius, tangent, and the line joining the centre to the external point. For example, if a tangent of length 24 cm is drawn from a point 25 cm away from the centre, the radius can be found as sqrt(625 - 576) = 7 cm. Similarly, if the radius is 5 cm and the external distance is 13 cm, the tangent length equals sqrt(169 - 25) = 12 cm. These straightforward applications of the Pythagoras theorem form the basis of many exam questions worth 1 to 2 marks.
Another important problem type involves finding the sides of a triangle circumscribing a circle (incircle). In Exercise 10.2 Q12, students encounter a triangle ABC with an inscribed circle of radius 4 cm, where BC is divided into segments of 8 cm and 6 cm by the point of contact. Using equal tangent lengths from each vertex and combining the incircle area formula (Area = r x s) with Heron's formula, students can determine that AB = 15 cm and AC = 13 cm. This problem type requires careful algebraic manipulation and is a favourite among examiners for 4 to 5 mark questions.
Proof-Based Questions and Exam Strategy
Proof-based problems form the most important category in this chapter and are almost guaranteed to appear in the CBSE board exam. Students must master the proof that tangent lengths from an external point are equal, as this is the most frequently asked long answer question. The proof uses RHS congruence: in the two right triangles formed by the radii, tangent segments, and the line from centre to external point, the hypotenuse is common and the radii are equal. Other critical proofs include showing that tangents at the ends of a diameter are parallel (using alternate interior angles), that a parallelogram circumscribing a circle is a rhombus, and that opposite sides of a circumscribed quadrilateral subtend supplementary angles at the centre. When writing proofs, students should always state the Given information, what needs To Be Proved, any Construction required, and then present the Proof with clear logical steps. CBSE examiners award marks for each step, so showing all intermediate reasoning is essential.
Common Problem Patterns in Board Exams
CBSE board exams frequently test the relationship angle APB + angle AOB = 180 degrees, where PA and PB are tangents from external point P. Students should be comfortable finding either angle when the other is given. A common variation asks students to prove that angle PTQ = 2 angle OPQ for tangents from an external point, which requires combining the isosceles triangle property (TP = TQ) with the tangent-radius perpendicularity. Another pattern involves concentric circles, where a chord of the larger circle serves as a tangent to the smaller circle. In such problems, the perpendicular from the common centre to the chord equals the radius of the smaller circle, and the Pythagoras theorem gives half the chord length. The incircle of a right triangle is also tested frequently, where the inradius can be found using the formula r = (a + b - c)/2, with c being the hypotenuse.
Key Theorems and Results Summary
To summarize the essential results of Chapter 10: (1) A tangent to a circle is perpendicular to the radius at the point of contact. (2) Tangent lengths from an external point are equal. (3) The angle between tangents from an external point is supplementary to the angle at the centre. (4) For a circumscribed quadrilateral, opposite sides sum to equal values: AB + CD = AD + BC. (5) A parallelogram circumscribing a circle is always a rhombus. (6) Tangents at the ends of a diameter are parallel. (7) Opposite sides of a circumscribed quadrilateral subtend supplementary angles at the centre. (8) From a point inside a circle, zero tangents can be drawn; from a point on the circle, exactly one; and from an external point, exactly two. These results form the complete toolkit needed to solve all CBSE board exam questions from this chapter. Students should prioritize understanding the proofs, practising numerical problems involving Pythagoras theorem, and mastering the circumscribed polygon properties to score full marks in the board examination.
Frequently Asked Questions
How many exercises are there in NCERT Class 10 Maths Chapter 10 Circles for the 2026-27 syllabus?
There are 2 exercises in Chapter 10 Circles as per the CBSE 2026-27 syllabus. Exercise 10.1 has 4 questions covering basic concepts of tangents, secants, and the point of contact. Exercise 10.2 has 13 questions involving proofs of tangent theorems, numerical problems using Pythagoras theorem, and properties of circumscribed polygons.
What are the two main theorems in Chapter 10 Circles?
The two main theorems are: (1) Theorem 10.1 - The tangent at any point of a circle is perpendicular to the radius through the point of contact. (2) Theorem 10.2 - The lengths of tangents drawn from an external point to a circle are equal. Both theorems are important for the board exam, and students should memorize their proofs.
How many marks does Chapter 10 Circles carry in the CBSE Class 10 board exam?
Chapter 10 Circles typically carries 4 to 6 marks in the CBSE Class 10 board exam. It falls under Unit IV: Geometry, which has a total weightage of 15 marks. Questions usually appear as 1 MCQ (1 mark) and 1 short or long answer question (3 to 5 marks) involving tangent proofs or numerical problems.
What is the difference between a tangent and a secant?
A tangent is a line that touches a circle at exactly one point (called the point of contact), while a secant is a line that intersects a circle at two distinct points. A tangent just grazes the circle's surface, whereas a secant cuts through it. A tangent is perpendicular to the radius at the point of contact, but a secant has no such perpendicularity condition.
How do you find the length of a tangent from an external point?
If the radius of the circle is r, and the distance from the external point to the centre is d, then the length of the tangent is sqrt(d^2 - r^2). This uses the Pythagoras theorem in the right triangle formed by the radius, the tangent, and the line joining the centre to the external point. The right angle is at the point of contact (tangent perpendicular to radius).
What is the property of a quadrilateral circumscribing a circle?
If a quadrilateral ABCD circumscribes a circle (the circle touches all four sides), then the sum of the opposite sides are equal: AB + CD = AD + BC. This property is derived from the fact that tangent lengths from each vertex to the circle are equal. This property is frequently tested in board exams.
Why is a parallelogram circumscribing a circle always a rhombus?
For any circumscribed quadrilateral, AB + CD = AD + BC. In a parallelogram, opposite sides are equal: AB = CD and AD = BC. Substituting these into the circumscribed property gives 2AB = 2AD, so AB = AD. Since all four sides become equal (AB = BC = CD = DA), the parallelogram is a rhombus.
What are the most important questions from Chapter 10 for the CBSE board exam?
The most important questions are: (1) Prove that tangent lengths from an external point are equal (Theorem 10.2 proof), (2) Prove that a parallelogram circumscribing a circle is a rhombus (Ex 10.2 Q11), (3) Prove that opposite sides of a circumscribed quadrilateral subtend supplementary angles at the centre (Ex 10.2 Q13), (4) Find the sides of a triangle circumscribing an incircle (Ex 10.2 Q12), and (5) Prove angle PTQ = 2 angle OPQ for tangents from external point (PYQ).