Solved Examples

How to Use This Section

This is your practice powerhouse for Circles. The problems below go from one-step warm-ups to full board proofs. Nearly every one uses one of just three facts: (1) tangent ⊥ radius, (2) tangent length $=\sqrt{d^2-r^2}$, (3) tangents from an external point are equal. Keep these in view and work each solution with a pen.

Example 1: Tangent length

A point is 17 cm from the centre of a circle of radius 8 cm. Find the length of the tangent from the point.

Solution: $\sqrt{d^2-r^2}=\sqrt{17^2-8^2}=\sqrt{289-64}=\sqrt{225}=15$ cm.

Final Answer: 15 cm.

Example 2: Find the radius

The tangent from a point 10 cm from the centre of a circle is 6 cm long. Find the radius.

Solution: $r=\sqrt{d^2-\text{tangent}^2}=\sqrt{10^2-6^2}=\sqrt{64}=8$ cm.

Final Answer: 8 cm.

Example 3: Distance to the centre

The radius of a circle is 7 cm and the length of a tangent from a point $P$ is 24 cm. Find $OP$.

Solution: $OP=\sqrt{r^2+\text{tangent}^2}=\sqrt{7^2+24^2}=\sqrt{625}=25$ cm.

Final Answer: 25 cm.

Example 4: Angle with the radius

A tangent touches a circle at $P$. What is the angle between the tangent and the radius $OP$?

Solution: By Theorem 10.1 it is $90^\circ$.

Final Answer: $90^\circ$.

Example 5: Angle between tangents

Tangents $TP$, $TQ$ from $T$ touch a circle centre $O$ with $\angle POQ = 130^\circ$. Find $\angle PTQ$.

Solution: $\angle PTQ = 180^\circ - 130^\circ = 50^\circ$.

Final Answer: $50^\circ$.

Example 6: Central angle from tangent angle

Two tangents from $P$ are inclined at $60^\circ$. Find the angle they subtend at the centre.

Solution: $180^\circ - 60^\circ = 120^\circ$.

Final Answer: $120^\circ$.

Example 7: Half-angle at the centre

Tangents $PA,PB$ from $P$ to a circle centre $O$ are inclined at $80^\circ$. Find $\angle POA$.

Solution: $\angle AOB = 180^\circ-80^\circ=100^\circ$; $OP$ bisects it, so $\angle POA = 50^\circ$.

Final Answer: $50^\circ$.

Example 8: Equal tangents

Tangents from an external point $P$ touch a circle at $A$ and $B$. If $PA = 3x-2$ and $PB = x+6$ (in cm), find $PA$.

Solution: Equal tangents: $3x-2 = x+6 \Rightarrow 2x = 8 \Rightarrow x = 4$. So $PA = 3(4)-2 = 10$ cm.

Final Answer: $PA = 10$ cm.

Example 9: Concentric chord

Two concentric circles have radii 25 cm and 24 cm. Find the length of the chord of the bigger circle that touches the smaller.

Solution: $2\sqrt{25^2-24^2}=2\sqrt{625-576}=2\sqrt{49}=2\times7=14$ cm.

Final Answer: 14 cm.

Example 10: Quadrilateral circumscribing

$ABCD$ circumscribes a circle with $AB=5$, $BC=8$, $CD=9$. Find $AD$.

Solution: $AB+CD=AD+BC \Rightarrow 5+9=AD+8 \Rightarrow AD=6$ cm.

Final Answer: 6 cm.

Example 11: MCQ — tangent from Q

From a point $Q$, the tangent to a circle is 24 cm and $OQ = 25$ cm. The radius is: (A) 7 (B) 12 (C) 15 (D) 24.5 cm.

Solution: $r=\sqrt{25^2-24^2}=\sqrt{49}=7$ cm.

Final Answer: (A) 7 cm.

Example 12: MCQ — angle POQ = 110°

$TP,TQ$ are tangents to a circle centre $O$ with $\angle POQ = 110^\circ$. Then $\angle PTQ$ is: (A) 60° (B) 70° (C) 80° (D) 90°.

Solution: $\angle PTQ = 180^\circ-110^\circ = 70^\circ$.

Final Answer: (B) 70°.

Example 13: MCQ — tangents inclined at 80°

Tangents $PA,PB$ from $P$ (centre $O$) are inclined at $80^\circ$. Then $\angle POA$ is: (A) 50° (B) 60° (C) 70° (D) 80°.

Solution: $\angle AOB = 100^\circ$, bisected: $\angle POA = 50^\circ$.

Final Answer: (A) 50°.

Example 14: Perpendicular passes through centre

Prove that the perpendicular at the point of contact to a tangent to a circle passes through the centre.

Solution:

1. Let the tangent touch the circle at $P$; by Theorem 10.1, the radius $OP \perp$ tangent.
2. Through $P$ there is only one line perpendicular to the tangent, and it is $OP$.
3. So the perpendicular at $P$ is the line $OP$, which passes through the centre $O$. $\blacksquare$

Example 15: Chord bisected at contact

Prove that in two concentric circles, a chord of the larger circle that touches the smaller is bisected at the point of contact.

Solution:

1. The chord $AB$ touches the inner circle at $P$, so $OP \perp AB$ (tangent ⊥ radius).
2. A perpendicular from the centre to a chord bisects it, hence $AP = BP$. $\blacksquare$

Example 16: $\angle PTQ = 2\,\angle OPQ$

Tangents $TP,TQ$ are drawn from an external point $T$ to a circle centre $O$. Prove $\angle PTQ = 2\,\angle OPQ$.

Solution:

1. Let $\angle PTQ = \theta$. As $TP=TQ$, triangle $TPQ$ is isosceles: $\angle TPQ = 90^\circ - \tfrac{\theta}{2}$.
2. $\angle OPT = 90^\circ$ (radius ⊥ tangent).
3. $\angle OPQ = \angle OPT - \angle TPQ = 90^\circ-(90^\circ-\tfrac{\theta}{2}) = \tfrac{\theta}{2}$.
4. Hence $\angle PTQ = \theta = 2\,\angle OPQ$. $\blacksquare$

Example 17: Chord + two tangents (length TP)

$PQ$ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at $P$ and $Q$ meet at $T$. Find $TP$.

Solution:

1. $OT \perp PQ$ and bisects it, so $PR = 4$ cm (foot $R$). $OR = \sqrt{5^2-4^2}=3$ cm.
2. $\triangle TRP \sim \triangle PRO$ (AA): $\dfrac{TP}{PO} = \dfrac{RP}{RO} \Rightarrow \dfrac{TP}{5} = \dfrac{4}{3}$.
3. $TP = \dfrac{20}{3}$ cm.

Final Answer: $TP = \dfrac{20}{3}$ cm.

Example 18: Tangents at ends of a diameter are parallel

Prove it.

Solution: The tangent at each end is perpendicular to the radius there, and both radii lie along the diameter. Two lines perpendicular to the same line (the diameter) are parallel. $\blacksquare$

Example 19: $\angle AOB = 90^\circ$ for the transversal tangent

$XY$ and $X'Y'$ are parallel tangents to a circle centre $O$; a tangent $AB$ (contact $C$) meets them at $A$ and $B$. Prove $\angle AOB = 90^\circ$.

Solution:

1. $OA$ bisects $\angle XAB$ and $OB$ bisects $\angle X'BA$ (tangents from $A$, resp. $B$).
2. $\angle XAB + \angle X'BA = 180^\circ$ (co-interior, parallel lines).
3. So $\angle OAB + \angle OBA = \tfrac12(180^\circ) = 90^\circ$, giving $\angle AOB = 180^\circ - 90^\circ = 90^\circ$. $\blacksquare$

Example 20: Rhombus

Prove a parallelogram circumscribing a circle is a rhombus.

Solution: $AB+CD=AD+BC$; with $AB=CD$, $AD=BC$ this gives $AB=BC$, so all sides equal — a rhombus. $\blacksquare$

Example 21: Incircle — find the sides

A triangle $ABC$ circumscribes a circle of radius 4 cm; the contact point on $BC$ gives $BD=8$, $DC=6$. Find $AB, AC$.

Solution: With $AF=AE=x$: sides $14, 8+x, 6+x$, $s=14+x$. Area $=4(14+x)=\sqrt{(14+x)x\cdot48}$. Squaring: $16(14+x)=48x \Rightarrow x=7$. So $AB=15$ cm, $AC=13$ cm.

Final Answer: $AB=15$ cm, $AC=13$ cm.

Example 22: Angle between tangent and chord set-up

Two tangents from $T$ touch a circle at $P,Q$ and $\angle PTQ = 50^\circ$. Find $\angle OPQ$.

Solution: $\angle OPQ = \tfrac12\angle PTQ = 25^\circ$.

Final Answer: $25^\circ$.

Example 23: Supplementary angle

The tangents from an external point subtend $70^\circ$ at the centre. Find the angle between the tangents.

Solution: $180^\circ - 70^\circ = 110^\circ$.

Final Answer: $110^\circ$.

Example 24: Two circles, common external tangent (numeric)

The radius of a circle is 5 cm. A tangent from a point $P$ has length 12 cm. Find the distance of $P$ from the nearest point of the circle.

Solution: $OP=\sqrt{5^2+12^2}=13$ cm. Nearest point of circle is at distance $OP-r = 13-5 = 8$ cm.

Final Answer: 8 cm.

Example 25: Quadrilateral — opposite side sums

$ABCD$ circumscribes a circle with $AB=7$, $CD=4$. Find $AD+BC-AB-CD$… i.e. find $AD+BC$.

Solution: $AD+BC = AB+CD = 7+4 = 11$ cm.

Final Answer: $AD+BC = 11$ cm.

Example 26: Diameter and tangent

A tangent to a circle of radius 6 cm is drawn from a point 10 cm from the centre. How long is the tangent?

Solution: $\sqrt{10^2-6^2}=\sqrt{64}=8$ cm.

Final Answer: 8 cm.

Example 27: Two tangents form an equilateral triangle

From $T$, tangents $TP,TQ$ touch a circle. If $\angle PTQ = 60^\circ$, show $\triangle TPQ$ is equilateral.

Solution: $TP=TQ$ so base angles equal, each $=\tfrac12(180^\circ-60^\circ)=60^\circ$. All three angles $60^\circ$ $\Rightarrow$ equilateral. $\blacksquare$

Example 28: Find the inradius-independent side

$ABCD$ circumscribes a circle. $AB=12$, $BC=10$, $CD=8$. Find $DA$.

Solution: $DA = AB+CD-BC = 12+8-10 = 10$ cm.

Final Answer: 10 cm.

Example 29: Opposite sides subtend supplementary angles

State the result about opposite sides of a quadrilateral circumscribing a circle (angles at the centre).

Solution: The opposite sides subtend supplementary angles at the centre: $\angle AOB + \angle COD = 180^\circ$ and $\angle BOC + \angle DOA = 180^\circ$.

Final Answer: Opposite sides subtend supplementary angles at the centre.

Example 30: Mixed — full reasoning

Tangents $PA$ and $PB$ are drawn from an external point $P$ to a circle centre $O$, radius 5 cm, with $OP = 13$ cm. Find (i) $PA$, (ii) $\angle APB$ if $\angle AOB = 120^\circ$.

Solution: (i) $PA = \sqrt{OP^2 - OA^2} = \sqrt{13^2 - 5^2} = \sqrt{144} = 12$ cm. (ii) $\angle APB = 180^\circ - \angle AOB = 180^\circ - 120^\circ = 60^\circ$.

Final Answer: $PA = 12$ cm; $\angle APB = 60^\circ$.

Takeaway: One figure, two tools — Pythagoras for lengths, the supplementary relation for angles.