Class 10 Circles (Tangents) Practice Questions with Solutions

Short answer: one or two steps#

Ques­tion 1#

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

Let the tan­gent touch the cir­cle at $T$. The radius $OT$ is per­pen­dic­u­lar to the tan­gent, so $\triangle OTP$ is right-angled at $T$ with hypotenuse $OP$.

$$PT = \sqrt{OP^2 - OT^2} = \sqrt{289 - 64} = \sqrt{225} = 15 \text{ cm}$$

Answer: $15$ cm.

Ques­tion 2#

$PA$ and $PB$ are tan­gents from $P$ to a cir­cle with cen­tre $O$. If $\angle APB = 50^\circ$, find $\angle AOB$.

In quadri­lat­eral $OAPB$, $\angle OAP = \angle OBP = 90^\circ$, so

$$\angle AOB = 360^\circ - 90^\circ - 90^\circ - 50^\circ = 130^\circ.$$

Answer: $\angle AOB = 130^\circ$.

Ques­tion 3#

Two tan­gents $PA$ and $PB$, each of length $10$ cm, are drawn from $P$ to a cir­cle, and $\angle APB = 60^\circ$. Find the length of the chord $AB$.

$PA = PB$, so $\triangle PAB$ is isosce­les and $\displaystyle \angle PAB = \angle PBA = \frac{180^\circ - 60^\circ}{2} = 60^\circ$. All three angles are $60^\circ$, so the tri­an­gle is equi­lat­eral and $AB = PA$.

Answer: $AB = 10$ cm.

Ques­tion 4#

Two con­cen­tric cir­cles have radii $4$ cm and $5$ cm. Find the length of the chord of the larger cir­cle that touches the smaller cir­cle.

Let the chord $AB$ touch the smaller cir­cle at $M$. Then $OM \perp AB$, $OM = 4$ cm, and $M$ is the mid­point of $AB$ (the per­pen­dic­u­lar from the cen­tre bisects a chord). In right tri­an­gle $OMA$ with $OA = 5$ cm:

$$AM = \sqrt{25 - 16} = 3 \text{ cm}, \qquad AB = 2 \times 3 = 6 \text{ cm}.$$

Answer: $6$ cm.

Ques­tion 5#

A cir­cle has radius $4.5$ cm. Two par­al­lel lines are both tan­gents to it. How far apart are the lines?

Two par­al­lel tan­gents touch the cir­cle at the ends of a diam­e­ter: the radii to the two points of con­tact are both per­pen­dic­u­lar to the same direc­tion, so they lie along one line through the cen­tre. The dis­tance between the lines is there­fore the diam­e­ter.

$$2 \times 4.5 = 9 \text{ cm}$$

Answer: $9$ cm.

Ques­tion 6#

A quadri­lat­eral $ABCD$ is drawn to cir­cum­scribe a cir­cle. If $AB = 6$ cm, $BC = 7$ cm and $CD = 4$ cm, find $AD$.

For a quadri­lat­eral cir­cum­scrib­ing a cir­cle, $AB + CD = AD + BC$.

$$6 + 4 = AD + 7 \;\Rightarrow\; AD = 3 \text{ cm}$$

Answer: $AD = 3$ cm.

Why this works: each side is made of two tan­gent pieces, and the four pairs of equal pieces are shared equally between $AB + CD$ and $AD + BC$.

Short answer: three-mark ques­tions#

Ques­tion 7#

$PA$ and $PB$ are tan­gents to a cir­cle with cen­tre $O$, and $\angle APB = 40^\circ$. Find $\angle OAB$.

Since $PA = PB$, tri­an­gle $PAB$ is isosce­les:

$$\displaystyle \angle PAB = \frac{180^\circ - 40^\circ}{2} = 70^\circ.$$

Since $OA \perp PA$, $\angle OAP = 90^\circ$. There­fore

$$\angle OAB = \angle OAP - \angle PAB = 90^\circ - 70^\circ = 20^\circ.$$

Answer: $\angle OAB = 20^\circ$.

Why this works: in gen­eral $\displaystyle \angle OAB = 90^\circ - \left(90^\circ - \tfrac{1}{2}\angle APB\right) = \tfrac{1}{2}\angle APB$.

Ques­tion 8#

The incir­cle of tri­an­gle $ABC$ touches $BC$, $CA$ and $AB$ at $D$, $E$ and $F$. If $AB = 12$ cm, $BC = 8$ cm and $CA = 10$ cm, find $AF$, $BD$ and $CE$.

Tan­gents from a ver­tex are equal. Let $AF = AE = x$, $BF = BD = y$ and $CD = CE = z$. Then

$$\begin{aligned} x + y &= 12 \\ y + z &= 8 \\ z + x &= 10 \end{aligned}$$

Adding: $2(x + y + z) = 30$, so $x + y + z = 15$. Sub­tract­ing each equa­tion from this:

$$x = 15 - 8 = 7, \qquad y = 15 - 10 = 5, \qquad z = 15 - 12 = 3.$$

Check: $7 + 5 = 12$, $5 + 3 = 8$, $3 + 7 = 10$.

Answer: $AF = 7$ cm, $BD = 5$ cm, $CE = 3$ cm.

Ques­tion 9#

A cir­cle is inscribed in a tri­an­gle $ABC$ right-angled at $B$, with $AB = 7$ cm and $BC = 24$ cm. Find the radius of the cir­cle.

By Pythago­ras, $AC = \sqrt{49 + 576} = \sqrt{625} = 25$ cm. Let the cir­cle, with cen­tre $O$ and radius $r$, touch $AB$, $BC$, $CA$ at $F$, $D$, $E$.

1. $OF \perp AB$, $OD \perp BC$ and $\angle B = 90^\circ$, so $OFBD$ is a rec­tan­gle with $OF = OD = r$, that is, a square. So $BF = BD = r$.
2. Then $AE = AF = 7 - r$ and $CE = CD = 24 - r$.
3. $AE + CE = AC$ gives $(7 - r) + (24 - r) = 25$.

$$31 - 2r = 25 \;\Rightarrow\; r = 3 \text{ cm}$$

Check with areas: area $\displaystyle = \tfrac{1}{2} \times 7 \times 24 = 84$ and half-perime­ter $= 28$; $84 \div 28 = 3$.

Answer: $3$ cm.

Ques­tion 10#

From a point $P$, $26$ cm from the cen­tre $O$ of a cir­cle of radius $10$ cm, tan­gents $PA$ and $PB$ are drawn. Find the length of each tan­gent and the length of the chord $AB$.

In right tri­an­gle $OAP$: $PA = \sqrt{26^2 - 10^2} = \sqrt{676 - 100} = \sqrt{576} = 24$ cm, and $PB = PA = 24$ cm.

$OP$ is the per­pen­dic­u­lar bisec­tor of $AB$. Let it meet $AB$ at $M$, so $AM \perp OP$ and $AB = 2AM$. Write the area of $\triangle OAP$ in two ways:

$$\displaystyle \begin{aligned} \tfrac{1}{2} \times OA \times AP &= \tfrac{1}{2} \times OP \times AM \\ 10 \times 24 &= 26 \times AM \\ AM &= \frac{240}{26} = \frac{120}{13} \text{ cm} \end{aligned}$$

$$\displaystyle AB = 2 \times \frac{120}{13} = \frac{240}{13} \text{ cm} \approx 18.46 \text{ cm}$$

Answer: $PA = PB = 24$ cm; $\displaystyle AB = \frac{240}{13}$ cm.

Ques­tion 11#

A cir­cle touches all four sides of a par­al­lel­o­gram $ABCD$. Show that $ABCD$ must have all its sides equal, that is, it is a rhom­bus.

Let the par­al­lel­o­gram $ABCD$ cir­cum­scribe a cir­cle, with $AB$, $BC$, $CD$, $DA$ touch­ing it at $P$, $Q$, $R$, $S$.

1. Tan­gents from an exter­nal point are equal: $AP = AS$, $BP = BQ$, $CR = CQ$, $DR = DS$.
2. Adding: $(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$, so $AB + CD = AD + BC$.
3. In a par­al­lel­o­gram, oppo­site sides are equal: $CD = AB$ and $AD = BC$.
4. Sub­sti­tut­ing: $2AB = 2BC$, so $AB = BC$.
5. Then $AB = BC = CD = DA$, so all four sides are equal.

Answer: A par­al­lel­o­gram with all sides equal is a rhom­bus, which proves the result. $\blacksquare$

Long answer: proofs and exam-level prob­lems#

Ques­tion 12#

A cir­cle with cen­tre $O$ lies between two par­al­lel lines $XY$ and $X'Y'$ and touches them at $P$ and $Q$. A third line touches the cir­cle at $C$ and cuts $XY$ at $A$ and $X'Y'$ at $B$. Show that $OA$ and $OB$ are per­pen­dic­u­lar.

Proof. Join $OP$, $OQ$, $OC$, $OA$ and $OB$.

1. In $\triangle OPA$ and $\triangle OCA$: $\angle OPA = \angle OCA = 90^\circ$ (radius $\perp$ tan­gent), $OA = OA$ (com­mon) and $OP = OC$ (radii). So $\triangle OPA \cong \triangle OCA$ (RHS), and $\angle OAP = \angle OAC$ (CPCT). Hence $\displaystyle \angle OAC = \tfrac{1}{2}\angle PAB$.
2. In the same way, $\triangle OQB \cong \triangle OCB$, so $\displaystyle \angle OBC = \tfrac{1}{2}\angle QBA$.
3. $XY \parallel X'Y'$ and $AB$ is a trans­ver­sal, so the co-inte­rior angles are sup­ple­men­tary: $\angle PAB + \angle QBA = 180^\circ$.
4. There­fore $\displaystyle \angle OAB + \angle OBA = \tfrac{1}{2}(\angle PAB + \angle QBA) = \tfrac{1}{2} \times 180^\circ = 90^\circ$.
5. In $\triangle AOB$: $\angle AOB = 180^\circ - (\angle OAB + \angle OBA) = 180^\circ - 90^\circ = 90^\circ$.

Answer: $\angle AOB = 90^\circ$, as required. $\blacksquare$

Ques­tion 13#

All four sides of a quadri­lat­eral $ABCD$ touch a cir­cle with cen­tre $O$. Show that $\angle AOB + \angle COD = 180^\circ$ and $\angle BOC + \angle AOD = 180^\circ$.

Proof. Let $ABCD$ cir­cum­scribe a cir­cle with cen­tre $O$, touch­ing $AB$, $BC$, $CD$, $DA$ at $P$, $Q$, $R$, $S$. Join $O$ to the four ver­tices and the four points of con­tact.

1. $\triangle OAP \cong \triangle OAS$ by RHS (right angles at $P$ and $S$, com­mon hypotenuse $OA$, radii $OP = OS$). So $\angle AOP = \angle AOS$; call each angle $a$.
2. Sim­i­larly $\angle BOP = \angle BOQ = b$, $\angle COQ = \angle COR = c$ and $\angle DOR = \angle DOS = d$.
3. These eight angles fill the com­plete angle round $O$: $2a + 2b + 2c + 2d = 360^\circ$, so $a + b + c + d = 180^\circ$.
4. Now $\angle AOB = a + b$ and $\angle COD = c + d$.

$$\angle AOB + \angle COD = a + b + c + d = 180^\circ$$

In the same way, $\angle BOC + \angle AOD = (b + c) + (d + a) = 180^\circ$.

Answer: Each pair of oppo­site sides sub­tends sup­ple­men­tary angles at the cen­tre. $\blacksquare$

Ques­tion 14#

A cir­cle of radius $3.5$ cm is inscribed in tri­an­gle $ABC$ and touches $BC$ at $D$, with $BD = 7$ cm and $DC = 14$ cm. Find $AB$ and $AC$.

Let the cir­cle touch $CA$ at $E$ and $AB$ at $F$, and let $AF = AE = x$. Tan­gents from $B$ and $C$ give $BF = BD = 7$ and $CE = CD = 14$. So

$$AB = x + 7, \qquad AC = x + 14, \qquad BC = 21.$$

Area using the radius. Join­ing $O$ to the ver­tices splits the tri­an­gle into three tri­an­gles, each with height $r$ on one side:

$$\displaystyle \text{area} = \tfrac{1}{2} r (AB + BC + CA) = r s,$$

where $\displaystyle s = \tfrac{1}{2}(x + 7 + 21 + x + 14) = x + 21$. So area $= 3.5(x + 21)$.

Area using Heron's for­mula. $s - BC = x$, $s - CA = 7$, $s - AB = 14$, so

$$\text{area} = \sqrt{(x + 21) \times x \times 7 \times 14} = \sqrt{98x(x + 21)}.$$

Equat­ing and squar­ing:

$$\begin{aligned} 98x(x + 21) &= 12.25(x + 21)^2 \\ 98x &= 12.25(x + 21) \quad (\text{dividing by } x + 21 \gt 0) \\ 8x &= x + 21 \\ x &= 3 \end{aligned}$$

So $AB = 3 + 7 = 10$ cm and $AC = 3 + 14 = 17$ cm. Check: $s = 24$, area $= \sqrt{24 \times 3 \times 7 \times 14} = \sqrt{7056} = 84$, and $rs = 3.5 \times 24 = 84$.

Answer: $AB = 10$ cm and $AC = 17$ cm.

Ques­tion 15#

$PQ$ is a chord of length $12$ cm of a cir­cle of radius $10$ cm with cen­tre $O$. The tan­gents at $P$ and $Q$ meet at $T$. Find the length $TP$.

Join $OT$, meet­ing $PQ$ at $R$.

1. $TP = TQ$ (tan­gents from $T$) and $OP = OQ$ (radii), so both $O$ and $T$ are equidis­tant from $P$ and $Q$. Hence $OT$ is the per­pen­dic­u­lar bisec­tor of $PQ$: $\angle ORP = 90^\circ$ and $PR = 6$ cm.
2. In right tri­an­gle $ORP$: $OR = \sqrt{10^2 - 6^2} = 8$ cm.
3. Let $TR = y$. In right tri­an­gle $TRP$: $TP^2 = y^2 + 36$.
4. In right tri­an­gle $OPT$ (right-angled at $P$, since $OP \perp TP$): $TP^2 = OT^2 - OP^2 = (y + 8)^2 - 100$.

$$\begin{aligned} y^2 + 36 &= y^2 + 16y + 64 - 100 \\ 36 &= 16y - 36 \\ y &= 4.5 \end{aligned}$$

$$TP = \sqrt{4.5^2 + 6^2} = \sqrt{20.25 + 36} = \sqrt{56.25} = 7.5 \text{ cm}$$

Check: $OT = 12.5$ cm and $\sqrt{12.5^2 - 10^2} = \sqrt{56.25} = 7.5$ cm.

Answer: $TP = 7.5$ cm.

Ques­tion 16#

A quadri­lat­eral $ABCD$ with $\angle D = 90^\circ$ cir­cum­scribes a cir­cle with cen­tre $O$. The cir­cle touches $AB$, $BC$, $CD$ and $DA$ at $P$, $Q$, $R$ and $S$. If $BC = 30$ cm, $BQ = 22$ cm and $CD = 18$ cm, find the radius of the cir­cle.

1. $CQ = BC - BQ = 30 - 22 = 8$ cm.
2. Tan­gents from $C$ are equal: $CR = CQ = 8$ cm.
3. $DR = CD - CR = 18 - 8 = 10$ cm.
4. In quadri­lat­eral $ORDS$: $\angle ORD = 90^\circ$ and $\angle OSD = 90^\circ$ (radius $\perp$ tan­gent), and $\angle D = 90^\circ$. So $ORDS$ is a rec­tan­gle. Its adja­cent sides $OR$ and $OS$ are radii, so they are equal, and $ORDS$ is a square.
5. Hence the radius $OR = DR = 10$ cm.

Answer: The radius is $10$ cm.

Why this works: a right angle at a ver­tex of a cir­cum­scribed fig­ure always cre­ates a small square whose side is the radius.

Key terms#

Tan­gent

A line that meets a cir­cle at exactly one point.

Point of con­tact

The point where a tan­gent touches the cir­cle.

Length of a tan­gent

The dis­tance from an exter­nal point to the point of con­tact.

Con­cen­tric cir­cles

Cir­cles with a com­mon cen­tre.

Cir­cum­scribed quadri­lat­eral

A quadri­lat­eral whose four sides all touch one cir­cle.

Incir­cle

The cir­cle inside a tri­an­gle touch­ing all three sides; its radius is the inra­dius.

Semi-perime­ter

Half the perime­ter of a tri­an­gle, writ­ten $s$.

RHS con­gru­ence

Right angle, hypotenuse and one side equal: the rule used to prove tan­gent results.

Com­mon ques­tions#

Which results can I quote with­out proof in the exam?#

The two the­o­rems of the chap­ter, that the tan­gent is per­pen­dic­u­lar to the radius and that tan­gents from an exter­nal point are equal, can be quoted with their names as rea­sons. If a ques­tion asks you to prove one of them, write the full proof.

How do I recog­nise a con­cen­tric-cir­cle ques­tion?#

Look for "a chord of the larger cir­cle which touches the smaller cir­cle". The point of con­tact is the foot of the per­pen­dic­u­lar from the cen­tre, so it is also the mid­point of the chord.

What is the fastest way to han­dle an incir­cle prob­lem?#

Label the two equal tan­gent pieces from each ver­tex with one let­ter, write each side as a sum, and solve. If the radius is involved, use area $= rs$, together with Heron's for­mula or the right-angle area for­mula.

Why does a right angle at a ver­tex give a square?#

The two radii to the points of con­tact are per­pen­dic­u­lar to the two sides, and the ver­tex angle is $90^\circ$. That makes four right angles with two adja­cent sides equal to the radius, which is a square.

How many marks do these ques­tions usu­ally carry?#

Sin­gle-step cal­cu­la­tions are usu­ally 1–2 marks; prob­lems that com­bine two results, such as Ques­tions 7 to 11, are typ­i­cally 3 marks; and proofs or multi-step prob­lems like Ques­tions 12 to 16 can be 4–5 marks.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.
3. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.
4. Kise­lev, A. P. Kise­lev's Geom­e­try, Book I: Planime­try (adapted by A. Given­tal). Sum­iz­dat.