This page provides comprehensive Circles – Exercise 10.2. Get step-by-step NCERT solutions for Class 10 Maths Chapter 10 Circles Exercise 10.2. Master Tangent Properties and Proofs.
Tangent Properties: Proofs and Applications
Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal.
Property: Tangents from an external point subtend equal angles at the center.
1. From a point Q, the length of the tangent to a circle is 12 cm and the distance of Q from the centre is 13 cm. The radius of the circle is:
(A) 7 cm (B) 12 cm (C) 5 cm (D) 25 cm
Tangent is perpendicular to radius ($\angle OPQ = 90^\circ$).
In right $\triangle OPQ$: $OQ^2 = OP^2 + PQ^2$
$13^2 = r^2 + 12^2 \Rightarrow 169 = r^2 + 144$
$r^2 = 25 \Rightarrow r = 5$ cm.
Option (C) is correct.
2. In the figure, if TP and TQ are the two tangents to a circle with centre O so that $\angle POQ = 120^\circ$, then $\angle PTQ$ is equal to:
(A) $60^\circ$ (B) $70^\circ$ (C) $80^\circ$ (D) $90^\circ$
In quadrilateral POQT, $\angle P = 90^\circ$ and $\angle Q = 90^\circ$.
Sum of angles = $360^\circ$.
$\angle PTQ + \angle POQ = 180^\circ$
$\angle PTQ + 120^\circ = 180^\circ \Rightarrow \angle PTQ = 60^\circ$.
Option (A) is correct.
3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of $60^\circ$, then $\angle POA$ is equal to:
(A) $50^\circ$ (B) $60^\circ$ (C) $70^\circ$ (D) $80^\circ$
$\triangle OAP \cong \triangle OBP$ (RHS).
So, $\angle APO = \angle BPO = \frac{1}{2} \angle APB = 30^\circ$.
In $\triangle OAP$: $\angle OAP = 90^\circ$.
$\angle POA = 180^\circ - (90^\circ + 30^\circ) = 60^\circ$.
Option (B) is correct.
4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Let PQ be a diameter. Let tangents AB and CD be drawn at P and Q respectively.
Since radius $\perp$ tangent, $\angle OPB = 90^\circ$ and $\angle OQC = 90^\circ$.
Consider transversal PQ. Alternate interior angles $\angle APQ$ and $\angle PQD$ are both $90^\circ$.
Since alternate angles are equal, lines AB and CD are parallel.
5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Let tangent be AB at point P. We know radius OP $\perp$ AB, so $\angle OPA = 90^\circ$.
If there is another point O' such that O'P $\perp$ AB, then $\angle O'PA = 90^\circ$.
This implies $\angle OPA = \angle O'PA$, which is only possible if O and O' lie on the same line perpendicular to the tangent.
Thus, the perpendicular must pass through the centre.
6. The length of a tangent from a point A at distance 10 cm from the centre of the circle is 8 cm. Find the radius of the circle.
Let O be center, T be point of contact. $OA = 10$ cm, $AT = 8$ cm.
In right $\triangle OTA$: $OT^2 + AT^2 = OA^2$
$r^2 + 8^2 = 10^2 \Rightarrow r^2 = 100 - 64 = 36$
$r = 6$ cm.
7. Two concentric circles are of radii 13 cm and 5 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Let chord be AB touching smaller circle at P. OP $\perp$ AB.
In right $\triangle APO$: $AO^2 = AP^2 + OP^2$
$13^2 = AP^2 + 5^2 \Rightarrow 169 = AP^2 + 25$
$AP^2 = 144 \Rightarrow AP = 12$ cm.
Total length $AB = 2 \times AP = 24$ cm.
8. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that $AB + CD = AD + BC$.
Tangents from external point are equal:
$AP = AS, BP = BQ, CR = CQ, DR = DS$.
Add LHS and RHS:
$(AP+BP) + (CR+DR) = (AS+DS) + (BQ+CQ)$
$AB + CD = AD + BC$.
Hence Proved.
9. In a circle with centre O, XY and X'Y' are two parallel tangents, and another tangent AB with point of contact C intersects XY at A and X'Y' at B. Prove that $\angle AOB = 90^\circ$.
Join OC.
$\triangle OPA \cong \triangle OCA$ (RHS) $\Rightarrow \angle POA = \angle COA$.
Similarly $\angle QOB = \angle COB$.
POQ is a diameter, so sum of angles on line is $180^\circ$.
$2\angle COA + 2\angle COB = 180^\circ \Rightarrow 2(\angle COA + \angle COB) = 180^\circ$.
$\angle AOB = 90^\circ$.
10. 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 centre.
In quadrilateral OAPB (where P is external point), angles at contact points A and B are $90^\circ$.
Sum of angles = $360^\circ$.
$\angle APB + \angle AOB + 90 + 90 = 360$
$\angle APB + \angle AOB = 180^\circ$.
Hence angles are supplementary.
11. Prove that the parallelogram circumscribing a circle is a rhombus.
Let ABCD be the parallelogram. From Q8, $AB + CD = AD + BC$.
Since ABCD is a parallelogram, opposite sides are equal ($AB=CD, AD=BC$).
$2AB = 2AD \Rightarrow AB = AD$.
Since adjacent sides are equal, ABCD is a rhombus.
12. A triangle ABC is drawn to circumscribe a circle of radius 2 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 4 cm and 3 cm respectively. Find the sides AB and AC.
$BD=4, DC=3$. Tangents equal: $BE=4, CF=3$. Let $AE=AF=x$.
Sides: $a=7, b=3+x, c=4+x$. Semi-perimeter $s = 7+x$.
Area $\triangle ABC = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(7+x)(x)(4)(3)} = \sqrt{12x(7+x)}$.
Area also = $\frac{1}{2}r(P) = \frac{1}{2}(2)(14+2x) = 14+2x$.
Square both sides: $12x(7+x) = 4(7+x)^2 \Rightarrow 3x = 7+x \Rightarrow x = 3.5$.
$AB = 4 + 3.5 = 7.5$ cm, $AC = 3 + 3.5 = 6.5$ cm.
13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Join center O to all points of contact and vertices.
We get pairs of congruent triangles (e.g., $\triangle OAP \cong \triangle OAS$).
Adjacent angles at center are equal (say a, a, b, b, c, c, d, d).
$2(a+b+c+d) = 360^\circ \Rightarrow a+b+c+d = 180^\circ$.
Angle by AB ($a+b$) + Angle by CD ($c+d$) = $180^\circ$.
Hence Proved.