Tangent Properties PYQs
Class 10 Previous Year Questions (2014 – 2026)
Topic Overview
Theorem 10.2: "The lengths of tangents drawn from an external point to a circle are equal." This single theorem is the backbone of all complex Chapter 10 proofs, including inscribed triangles and circumscribed quadrilaterals.
Q1
2024
00:00
Prove that the lengths of tangents drawn from an external point to a circle are equal.
Given: A circle C(O, r) and two tangents PA, PB from external point P.
To Prove: $PA = PB$.
Construction: Join OA, OB, OP.
Proof: In $\triangle OAP$ and $\triangle OBP$:
1. $OA = OB$ (Radii)
2. $\angle OAP = \angle OBP = 90^{\circ}$ (Radius $\perp$ Tangent)
3. $OP = OP$ (Common)
By RHS congruence, $\triangle OAP \cong \triangle OBP$.
Thus $PA = PB$ (CPCT).
Ans: Proved
Q2
2023
00:00
A quadrilateral ABCD is drawn to circumscribe a circle. Prove that $AB + CD = AD + BC$.
Let P, Q, R, S be points of contact on sides AB, BC, CD, DA.
By Th 10.2: $AP = AS, BP = BQ, CR = CQ, DR = DS$.
Adding: $(AP+BP) + (CR+DR) = (AS+DS) + (BQ+CQ)$.
$AB + CD = AD + BC$.
Ans: Proved
Q3
2022
00:00
Prove 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 center.
In quadrilateral OAPB: $\angle OAP = 90^{\circ}, \angle OBP = 90^{\circ}$.
Sum of angles $= 360^{\circ}$.
$\angle APB + \angle AOB + 90 + 90 = 360$.
$\angle APB + \angle AOB = 180^{\circ}$.
Ans: Proved
Q4
2020
00:00
Prove that the parallelogram circumscribing a circle is a rhombus.
Let ABCD be a parallelogram circumscribing a circle.
From $AB+CD = AD+BC$ (Proved in Q2).
Since ABCD is a parallelogram, $AB=CD$ and $AD=BC$.
$2AB = 2AD \Rightarrow AB = AD$.
Since adjacent sides are equal, it is a rhombus.
Ans: Proved
Q5
2019
00:00
From a point P, two tangents PA and PB are drawn to a circle C(O, r). If $OP = 2r$, then find $\angle APB$. What type of triangle is APB?
In $\triangle OAP$: $\sin \angle OPA = OA/OP = r/2r = 1/2$.
$\angle OPA = 30^{\circ}$.
$\angle APB = 2 \times 30^{\circ} = 60^{\circ}$.
Since $PA=PB$ and $\angle P = 60^{\circ}$, $\triangle APB$ is equilateral.
Ans: 60°, Equilateral triangle
Q6
2024
00:00
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.
Let $AF = AE = x$. Sides are $14, x+6, x+8$.
Use Hero's formula for Area of $\triangle ABC$. $s = 14+x$.
Area $= \sqrt{(14+x)x \times 8 \times 6}$.
Also Area $= \text{Area}(\triangle OBC) + \text{Area}(\triangle OCA) + \text{Area}(\triangle OAB) = 1/2 \times r \times (14 + x+6 + x+8) = 2(28+2x) = 4(14+x)$.
Equating: $4(14+x) = \sqrt{48x(14+x)} \Rightarrow 16(14+x)^2 = 48x(14+x) \Rightarrow 14+x = 3x \Rightarrow x = 7$.
Sides: $AB = 7+8 = 15$ cm, $AC = 7+6 = 13$ cm.
Ans: AB=15 cm, AC=13 cm
Q7
2018
00:00
In the given figure, XY and X'Y' are two parallel tangents to a circle with center O and another tangent AB with point of contact C intersecting XY at A and X'Y' at B. Prove that $\angle AOB = 90^{\circ}$.
Join OC. In $\triangle OPA$ and $\triangle OCA$: $OP=OC, PA=AC, OA=OA$.
By SSS, $\triangle OPA \cong \triangle OCA \Rightarrow \angle 1 = \angle 2$.
Similarly, $\triangle OQB \cong \triangle OCB \Rightarrow \angle 4 = \angle 3$.
$"2(\angle 2 + \angle 3) = 180^{\circ}$ (angles on line POQ).
$\angle 2 + \angle 3 = 90^{\circ} \Rightarrow \angle AOB = 90^{\circ}$.
Ans: Proved
Q8
2017
00:00
In $\triangle ABC$ circumscribing a circle, sides $BC=12, CA=10, AB=8$. Find $AD, BE, CF$ where D, E, F are points of contact on sides AB, BC, CA.
Let $AD = AF = x, BE = BD = y, CF = CE = z$.
$x+y = 8, y+z = 12, z+x = 10$.
Adding all: $2(x+y+z) = 30 \Rightarrow x+y+z = 15$.
$x = 15 - 12 = 3$ cm.
$y = 15 - 10 = 5$ cm.
$z = 15 - 8 = 7$ cm.
Ans: AD=3cm, BE=5cm, CF=7cm
Q9
2018
00:00
If a hexagon ABCDEF circumscribes a circle, prove that $AB + CD + EF = BC + DE + FA$.
Let points of contact on AB, BC, CD, DE, EF, FA be P, Q, R, S, T, U.
$AP=AU, BP=BQ, CQ=CR, DR=DS, ES=ET, FT=FU$.
LHS $= (AP+BP) + (CR+DR) + (ET+FT)$.
RHS $= (BQ+CQ) + (DS+ES) + (FU+AU)$.
Substitute segments to show LHS = RHS.
Ans: Proved
Q10
2026 (SP)
00:00
PASSAGE: Designing a Logo. A sports club logo involves a circle inscribed in a quadrilateral. Side AB=7, BC=8, CD=9.
(i) Find the length of side AD.
(ii) If a tangent is drawn at point P, what is the angle between radius and tangent?
(iii) (a) If AB is parallel to CD, what type of quadrilateral is it? OR (b) State the theorem used to find AD.
(i) Find the length of side AD.
(ii) If a tangent is drawn at point P, what is the angle between radius and tangent?
(iii) (a) If AB is parallel to CD, what type of quadrilateral is it? OR (b) State the theorem used to find AD.
(i) $AB+CD = AD+BC \Rightarrow 7+9 = AD+8 \Rightarrow AD = 8$.
(ii) $90^{\circ}$.
(iii) (b) Theorem 10.2: Tangents from an external point are equal.
Prove that the lengths of tangents drawn from an external point to a circle are equal.
Given: A circle C(O, r) and two tangents PA, PB from external point P.
To Prove: $PA = PB$.
Construction: Join OA, OB, OP.
Proof: In $\triangle OAP$ and $\triangle OBP$:
1. $OA = OB$ (Radii)
2. $\angle OAP = \angle OBP = 90^{\circ}$ (Radius $\perp$ Tangent)
3. $OP = OP$ (Common)
By RHS congruence, $\triangle OAP \cong \triangle OBP$.
Thus $PA = PB$ (CPCT).
Ans: Proved
Q2
2023
00:00
A quadrilateral ABCD is drawn to circumscribe a circle. Prove that $AB + CD = AD + BC$.
Let P, Q, R, S be points of contact on sides AB, BC, CD, DA.
By Th 10.2: $AP = AS, BP = BQ, CR = CQ, DR = DS$.
Adding: $(AP+BP) + (CR+DR) = (AS+DS) + (BQ+CQ)$.
$AB + CD = AD + BC$.
Ans: Proved
Q3
2022
00:00
Prove 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 center.
In quadrilateral OAPB: $\angle OAP = 90^{\circ}, \angle OBP = 90^{\circ}$.
Sum of angles $= 360^{\circ}$.
$\angle APB + \angle AOB + 90 + 90 = 360$.
$\angle APB + \angle AOB = 180^{\circ}$.
Ans: Proved
Q4
2020
00:00
Prove that the parallelogram circumscribing a circle is a rhombus.
Let ABCD be a parallelogram circumscribing a circle.
From $AB+CD = AD+BC$ (Proved in Q2).
Since ABCD is a parallelogram, $AB=CD$ and $AD=BC$.
$2AB = 2AD \Rightarrow AB = AD$.
Since adjacent sides are equal, it is a rhombus.
Ans: Proved
Q5
2019
00:00
From a point P, two tangents PA and PB are drawn to a circle C(O, r). If $OP = 2r$, then find $\angle APB$. What type of triangle is APB?
In $\triangle OAP$: $\sin \angle OPA = OA/OP = r/2r = 1/2$.
$\angle OPA = 30^{\circ}$.
$\angle APB = 2 \times 30^{\circ} = 60^{\circ}$.
Since $PA=PB$ and $\angle P = 60^{\circ}$, $\triangle APB$ is equilateral.
Ans: 60°, Equilateral triangle
Q6
2024
00:00
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.
Let $AF = AE = x$. Sides are $14, x+6, x+8$.
Use Hero's formula for Area of $\triangle ABC$. $s = 14+x$.
Area $= \sqrt{(14+x)x \times 8 \times 6}$.
Also Area $= \text{Area}(\triangle OBC) + \text{Area}(\triangle OCA) + \text{Area}(\triangle OAB) = 1/2 \times r \times (14 + x+6 + x+8) = 2(28+2x) = 4(14+x)$.
Equating: $4(14+x) = \sqrt{48x(14+x)} \Rightarrow 16(14+x)^2 = 48x(14+x) \Rightarrow 14+x = 3x \Rightarrow x = 7$.
Sides: $AB = 7+8 = 15$ cm, $AC = 7+6 = 13$ cm.
Ans: AB=15 cm, AC=13 cm
Q7
2018
00:00
In the given figure, XY and X'Y' are two parallel tangents to a circle with center O and another tangent AB with point of contact C intersecting XY at A and X'Y' at B. Prove that $\angle AOB = 90^{\circ}$.
Join OC. In $\triangle OPA$ and $\triangle OCA$: $OP=OC, PA=AC, OA=OA$.
By SSS, $\triangle OPA \cong \triangle OCA \Rightarrow \angle 1 = \angle 2$.
Similarly, $\triangle OQB \cong \triangle OCB \Rightarrow \angle 4 = \angle 3$.
$"2(\angle 2 + \angle 3) = 180^{\circ}$ (angles on line POQ).
$\angle 2 + \angle 3 = 90^{\circ} \Rightarrow \angle AOB = 90^{\circ}$.
Ans: Proved
Q8
2017
00:00
In $\triangle ABC$ circumscribing a circle, sides $BC=12, CA=10, AB=8$. Find $AD, BE, CF$ where D, E, F are points of contact on sides AB, BC, CA.
Let $AD = AF = x, BE = BD = y, CF = CE = z$.
$x+y = 8, y+z = 12, z+x = 10$.
Adding all: $2(x+y+z) = 30 \Rightarrow x+y+z = 15$.
$x = 15 - 12 = 3$ cm.
$y = 15 - 10 = 5$ cm.
$z = 15 - 8 = 7$ cm.
Ans: AD=3cm, BE=5cm, CF=7cm
Q9
2018
00:00
If a hexagon ABCDEF circumscribes a circle, prove that $AB + CD + EF = BC + DE + FA$.
Let points of contact on AB, BC, CD, DE, EF, FA be P, Q, R, S, T, U.
$AP=AU, BP=BQ, CQ=CR, DR=DS, ES=ET, FT=FU$.
LHS $= (AP+BP) + (CR+DR) + (ET+FT)$.
RHS $= (BQ+CQ) + (DS+ES) + (FU+AU)$.
Substitute segments to show LHS = RHS.
Ans: Proved
Q10
2026 (SP)
00:00
PASSAGE: Designing a Logo. A sports club logo involves a circle inscribed in a quadrilateral. Side AB=7, BC=8, CD=9.
(i) Find the length of side AD.
(ii) If a tangent is drawn at point P, what is the angle between radius and tangent?
(iii) (a) If AB is parallel to CD, what type of quadrilateral is it? OR (b) State the theorem used to find AD.
(i) Find the length of side AD.
(ii) If a tangent is drawn at point P, what is the angle between radius and tangent?
(iii) (a) If AB is parallel to CD, what type of quadrilateral is it? OR (b) State the theorem used to find AD.
(i) $AB+CD = AD+BC \Rightarrow 7+9 = AD+8 \Rightarrow AD = 8$.
(ii) $90^{\circ}$.
(iii) (b) Theorem 10.2: Tangents from an external point are equal.