Tangents to a Circle PYQs
Class 10 Previous Year Questions (2014 – 2026)
Topic Overview
Master Theorem 10.1: "The tangent at any point of a circle is perpendicular to the radius through the point of contact." This fundamental property is used to solve distance problems using the Pythagoras theorem.
Q1
2024
00:00
If the radius of a circle is 5 cm and a tangent is drawn from a point 13 cm away from the center, find the length of the tangent.
Let $O$ be center, $P$ be external point, $Q$ be point of contact.
$OQ = 5$ (radius), $OP = 13$ (dist from center).
By Pythagoras theorem in $\triangle OQP$ ($OQ \perp QP$):
$QP^2 = OP^2 - OQ^2 = 13^2 - 5^2 = 169 - 25 = 144$.
$QP = \sqrt{144} = 12$ cm.
Ans: (A) 12 cm
Q2
2023
00:00
The distance between two parallel tangents of a circle of radius 4 cm is:
Parallel tangents are drawn at the ends of a diameter.
Distance $= \text{Diameter} = 2 \times \text{radius} = 2 \times 4 = 8$ cm.
Ans: (D) 8 cm
Q3
2020
00:00
If the angle between two radii of a circle is 130°, then find the angle between the tangents at the ends of the radii.
The angle between tangents and the angle between radii are supplementary.
$\text{Angle between tangents} = 180^{\circ} - 130^{\circ} = 50^{\circ}$.
Ans: 50°
Q4
2019
00:00
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the center O at a point Q so that OQ = 12 cm. Length PQ is:
$PQ^2 = OQ^2 - OP^2 = 12^2 - 5^2 = 144 - 25 = 119$.
$PQ = \sqrt{119}$ cm.
Ans: (D) $\sqrt{119}$ cm
Q7
2018
00:00
PA and PB are tangents to a circle from point P. If $\angle OAB = 30^{\circ}$, find $\angle APB$.
In $\triangle OAB$, $OA=OB$ (radii) $\Rightarrow \angle OBA = \angle OAB = 30^{\circ}$.
$\angle AOB = 180 - (30+30) = 120^{\circ}$.
$\angle APB = 180 - \angle AOB = 180 - 120 = 60^{\circ}$.
Ans: 60°
Q8
2024
00:00
From external point P, tangents PA and PB are drawn. If $\angle PAB = 50^{\circ}$, find $\angle AOB$.
$\angle OAP = 90^{\circ} \Rightarrow \angle OAB = 90 - 50 = 40^{\circ}$.
$\angle OBA = \angle OAB = 40^{\circ}$.
$\angle AOB = 180 - (40+40) = 100^{\circ}$.
Ans: 100°
Q9
2025 (SP)
00:00
A circle is inscribed in $\triangle ABC$, right angled at C. $AC = 8$ cm, $BC = 6$ cm. Find the radius $r$ of the circle.
$AB = \sqrt{8^2 + 6^2} = 10$ cm.
In a right triangle, radius of incircle $r = (AC + BC - AB)/2$.
$r = (8 + 6 - 10)/2 = 4/2 = 2$ cm.
Ans: 2 cm
Q5
2022
00:00
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 chord be AB, touching smaller circle at P. $OP \perp AB$.
In $\triangle OAP$: $OA = 5$ (outer radius), $OP = 3$ (inner radius).
$AP^2 = 5^2 - 3^2 = 25 - 9 = 16 \Rightarrow AP = 4$ cm.
$AB = 2 \times AP = 8$ cm.
Ans: 8 cm
Q6
2018
00:00
Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Consider a circle with center O and tangent XY at P.
Take any point Q other than P on XY.
Since Q lies outside the circle, OQ > OP (radius).
OP is the shortest distance from O to XY.
Hence $OP \perp XY$.
Ans: Proved
If the radius of a circle is 5 cm and a tangent is drawn from a point 13 cm away from the center, find the length of the tangent.
Let $O$ be center, $P$ be external point, $Q$ be point of contact.
$OQ = 5$ (radius), $OP = 13$ (dist from center).
By Pythagoras theorem in $\triangle OQP$ ($OQ \perp QP$):
$QP^2 = OP^2 - OQ^2 = 13^2 - 5^2 = 169 - 25 = 144$.
$QP = \sqrt{144} = 12$ cm.
Ans: (A) 12 cm
Q2
2023
00:00
The distance between two parallel tangents of a circle of radius 4 cm is:
Parallel tangents are drawn at the ends of a diameter.
Distance $= \text{Diameter} = 2 \times \text{radius} = 2 \times 4 = 8$ cm.
Ans: (D) 8 cm
Q3
2020
00:00
If the angle between two radii of a circle is 130°, then find the angle between the tangents at the ends of the radii.
The angle between tangents and the angle between radii are supplementary.
$\text{Angle between tangents} = 180^{\circ} - 130^{\circ} = 50^{\circ}$.
Ans: 50°
Q4
2019
00:00
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the center O at a point Q so that OQ = 12 cm. Length PQ is:
$PQ^2 = OQ^2 - OP^2 = 12^2 - 5^2 = 144 - 25 = 119$.
$PQ = \sqrt{119}$ cm.
Ans: (D) $\sqrt{119}$ cm
Q7
2018
00:00
PA and PB are tangents to a circle from point P. If $\angle OAB = 30^{\circ}$, find $\angle APB$.
In $\triangle OAB$, $OA=OB$ (radii) $\Rightarrow \angle OBA = \angle OAB = 30^{\circ}$.
$\angle AOB = 180 - (30+30) = 120^{\circ}$.
$\angle APB = 180 - \angle AOB = 180 - 120 = 60^{\circ}$.
Ans: 60°
Q8
2024
00:00
From external point P, tangents PA and PB are drawn. If $\angle PAB = 50^{\circ}$, find $\angle AOB$.
$\angle OAP = 90^{\circ} \Rightarrow \angle OAB = 90 - 50 = 40^{\circ}$.
$\angle OBA = \angle OAB = 40^{\circ}$.
$\angle AOB = 180 - (40+40) = 100^{\circ}$.
Ans: 100°
Q9
2025 (SP)
00:00
A circle is inscribed in $\triangle ABC$, right angled at C. $AC = 8$ cm, $BC = 6$ cm. Find the radius $r$ of the circle.
$AB = \sqrt{8^2 + 6^2} = 10$ cm.
In a right triangle, radius of incircle $r = (AC + BC - AB)/2$.
$r = (8 + 6 - 10)/2 = 4/2 = 2$ cm.
Ans: 2 cm
Q5
2022
00:00
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 chord be AB, touching smaller circle at P. $OP \perp AB$.
In $\triangle OAP$: $OA = 5$ (outer radius), $OP = 3$ (inner radius).
$AP^2 = 5^2 - 3^2 = 25 - 9 = 16 \Rightarrow AP = 4$ cm.
$AB = 2 \times AP = 8$ cm.
Ans: 8 cm
Q6
2018
00:00
Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Consider a circle with center O and tangent XY at P.
Take any point Q other than P on XY.
Since Q lies outside the circle, OQ > OP (radius).
OP is the shortest distance from O to XY.
Hence $OP \perp XY$.
Ans: Proved