Class 10 Mathematics Chapter 10 Question Answer | Circles | English Medium

Class 10 Mathematics Chapter 10 Question Answer | Circles | English Medium | ASSEB

Welcome to HSLC Guru

Hello dear student, welcome to HSLC Guru! On this page you will find the complete English-medium solutions for ASSEB Class 10 Mathematics, Chapter 10 — Circles. In Class 9 you have already studied that a circle is the set of all points in a plane that are at a fixed distance from a fixed point in the plane, and you met the basic terms — chord, arc, segment, sector, secant — together with several theorems about chords and angles. In Class 10 we narrow our focus to one very special situation: the relationship between a circle and a straight line that just touches it. This chapter introduces the precise idea of a tangent, proves the two fundamental theorems about tangents (Theorem 10.1 and Theorem 10.2), and then applies them to a wide range of geometric problems. Each problem in Exercises 10.1 and 10.2 of the ASSEB textbook is solved here step by step, with figures, reasoning and final answers. By the end of this lesson you will be able to compute tangent lengths, prove tangent properties and tackle every HSLC pattern question on circles.

Summary

A circle is the locus of all points in a plane that are at a constant distance (the radius) from a fixed point (the centre). A chord is a line segment joining two points on the circle, the longest chord through the centre is the diameter, a secant is a line that meets the circle at two distinct points, and a tangent is a line that meets the circle at exactly one point — this single common point is called the point of contact. The chapter is built on two key theorems. Theorem 10.1 says that the tangent at any point of a circle is perpendicular to the radius drawn through the point of contact. Theorem 10.2 says that the lengths of two tangents drawn from an external point to a circle are equal. Together with the fact that from a point outside a circle exactly two tangents can be drawn (zero from an interior point, exactly one from a point on the circle), these results solve almost every problem about tangents. The tangent–radius perpendicularity converts every tangent problem into a right-triangle problem, so the Pythagoras Theorem from Class 9 reappears as the central computational tool of this chapter.

Recap from Class 9 — Lines and a Circle

Consider a fixed circle with centre $O$ and a moving straight line in the plane of the circle. Three different things can happen depending on the perpendicular distance $d$ of the line from $O$ compared with the radius $r$.

Position of the line	Distance from centre	Number of points common with the circle	Name of the line
Line lies wholly outside the circle	$d > r$	$0$	Non-intersecting line
Line just touches the circle	$d = r$	$1$	Tangent
Line cuts the circle	$d < r$	$2$	Secant

Thus a tangent is the limiting position of a secant when the two points of intersection move toward each other and finally coincide.

Tangent to a Circle

Definition. A tangent to a circle is a line that intersects the circle at exactly one point. This common point is called the point of contact, and the tangent is said to touch the circle at that point.

Important consequences that follow at once from the definition:

A circle has infinitely many tangents — one at each of its points.
From a point inside the circle no tangent can be drawn.
From a point on the circle exactly one tangent can be drawn (the tangent at that point).
From a point outside the circle exactly two tangents can be drawn.
A circle can have at most two parallel tangents, and they are the tangents at the two ends of any diameter.

The two tangents at the endpoints of a diameter are parallel because each is perpendicular to the same diameter (Theorem 10.1).

Theorem 10.1 — The Tangent–Radius Theorem

Statement. The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Given. A circle with centre $O$ and a tangent $XY$ to the circle touching it at the point $P$.

To prove. $OP \perp XY$.

Construction. Take any point $Q$ (other than $P$) on the tangent $XY$. Join $OQ$.

Proof. The point $Q$ lies on the tangent $XY$. Now $XY$ touches the circle only at $P$, so $Q$ does not lie on the circle — it must lie outside the circle. Therefore

$$OQ > OP \qquad\text{(any point outside the circle is farther from the centre than the radius).}$$

The same is true for every point of $XY$ other than $P$. Hence $OP$ is the shortest distance from the centre $O$ to the line $XY$.

But the shortest distance from a point to a line is the perpendicular distance. Therefore $OP$ must be perpendicular to $XY$, that is

$$OP \perp XY.$$

Hence proved. $\blacksquare$

Important remarks.

By Theorem 10.1, at any point of a circle there is one and only one tangent.
The line containing the radius through the point of contact is called the normal to the circle at that point. Thus the normal at any point of a circle passes through the centre.
Converse (also true): if a line through a point $P$ of a circle is perpendicular to the radius $OP$, then the line is a tangent at $P$.

Number of Tangents from a Point to a Circle

Let $P$ be any point in the plane of a circle. Three cases arise.

Position of $P$	Number of tangents from $P$	Reason
$P$ lies inside the circle	$0$	Every line through $P$ meets the circle at two points.
$P$ lies on the circle	$1$	The tangent at $P$ itself; this tangent is perpendicular to $OP$.
$P$ lies outside the circle	$2$	Two distinct tangent lines can be drawn from $P$ to the circle.

The two points $T_1$ and $T_2$ where the tangents from an external point $P$ touch the circle are called the points of contact, and the segments $PT_1$ and $PT_2$ are called the lengths of the tangents from $P$.

Theorem 10.2 — The Two-Tangent Theorem

Statement. The lengths of tangents drawn from an external point to a circle are equal.

Given. A circle with centre $O$ and an external point $P$. $PQ$ and $PR$ are two tangents from $P$ touching the circle at $Q$ and $R$ respectively.

To prove. $PQ = PR$.

Construction. Join $OQ$, $OR$ and $OP$.

Proof. By Theorem 10.1, the tangent is perpendicular to the radius at the point of contact, so

$$\angle OQP = \angle ORP = 90°.$$

Now in the two right-angled triangles $\triangle OQP$ and $\triangle ORP$:

$OQ = OR$ (radii of the same circle)
$OP = OP$ (common hypotenuse)
$\angle OQP = \angle ORP = 90°$

Therefore by the RHS congruence rule,

$$\triangle OQP \cong \triangle ORP.$$

By CPCT (corresponding parts of congruent triangles), $PQ = PR$. Hence proved. $\blacksquare$

Useful corollaries that follow from the same congruence:

$\angle OPQ = \angle OPR$ — i.e. the centre of the circle lies on the bisector of the angle between the two tangents.
$\angle QOP = \angle ROP$ — i.e. the line $OP$ bisects $\angle QOR$.
The quadrilateral $OQPR$ has $\angle Q + \angle R = 180°$, so $\angle QOR + \angle QPR = 180°$ (the angle at the centre and the angle between the tangents are supplementary).

Worked Examples (NCERT/ASSEB)

Example 1. Prove that in two concentric circles, the chord of the larger circle which touches the smaller circle is bisected at the point of contact.

Solution. Let the two concentric circles have common centre $O$, with $AB$ a chord of the larger circle that touches the smaller circle at $P$. We have to show $AP = PB$.

Since $AB$ is tangent to the smaller circle at $P$, by Theorem 10.1, $OP \perp AB$. Now $AB$ is a chord of the larger circle and $OP$ (drawn from the centre $O$) is perpendicular to this chord. By the perpendicular-from-centre-bisects-chord property (proved in Class 9),

$$AP = PB.$$

Hence the chord is bisected at the point of contact. $\blacksquare$

Example 2. Two tangents $TP$ and $TQ$ are drawn to a circle with centre $O$ from an external point $T$. Prove that $\angle PTQ = 2\angle OPQ$.

Solution. Let $\angle PTQ = \theta$. We must show $\angle OPQ = \theta/2$.

Since $TP = TQ$ (Theorem 10.2), $\triangle TPQ$ is isosceles, so

$$\angle TPQ = \angle TQP = \tfrac{1}{2}(180° – \theta) = 90° – \tfrac{\theta}{2}.$$

By Theorem 10.1, $OP \perp TP$, i.e. $\angle OPT = 90°$. Therefore

$$\angle OPQ = \angle OPT – \angle TPQ = 90° – \left(90° – \tfrac{\theta}{2}\right) = \tfrac{\theta}{2}.$$

Hence $\angle PTQ = 2\angle OPQ$. $\blacksquare$

Example 3. PQ is a chord of length $8$ cm of a circle of radius $5$ cm. The tangents at $P$ and $Q$ intersect at a point $T$. Find the length $TP$.

Solution. Let $OT$ meet $PQ$ at $M$. By the corollaries of Theorem 10.2, $OT$ is the perpendicular bisector of $PQ$, so $PM = MQ = 4$ cm and $\angle OMP = 90°$. In $\triangle OMP$,

$$OM = \sqrt{OP^2 – PM^2} = \sqrt{25 – 16} = 3\text{ cm}.$$

Let $TP = x$ and $TM = y$. From $\triangle TMP$ (right-angled at $M$): $x^2 = y^2 + 16$. From $\triangle OPT$ (right-angled at $P$ by Theorem 10.1): $TP^2 + OP^2 = OT^2$, i.e. $x^2 + 25 = (y+3)^2$.

Subtracting: $25 = (y+3)^2 – y^2 – 16 = 6y – 7$, so $y = \dfrac{32}{6} = \dfrac{16}{3}$. Then

$$x^2 = \left(\tfrac{16}{3}\right)^2 + 16 = \tfrac{256}{9} + \tfrac{144}{9} = \tfrac{400}{9}, \quad x = \tfrac{20}{3}\text{ cm}.$$

Hence $TP = \dfrac{20}{3}$ cm.

Exercise 10.1 — Solutions

Q1. How many tangents can a circle have?

Answer: A circle has infinitely many tangents. At each point of the circle there is exactly one tangent (perpendicular to the radius at that point), and the circle has infinitely many points; therefore the total number of tangents is infinite.

Q2. Fill in the blanks:

(i) A tangent to a circle intersects it in ______ point(s).
(ii) A line intersecting a circle in two points is called a ______.
(iii) A circle can have ______ parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called ______.

Answer:

(i) one
(ii) secant
(iii) two (the tangents at the two ends of any diameter are parallel; no three tangents to a single circle can all be parallel)
(iv) point of contact

Q3. A tangent $PQ$ at a point $P$ of a circle of radius $5$ cm meets a line through the centre $O$ at a point $Q$ so that $OQ = 12$ cm. Length $PQ$ is: (A) $12$ cm (B) $13$ cm (C) $8.5$ cm (D) $\sqrt{119}$ cm.

Answer: Since $PQ$ is the tangent at $P$ and $OP$ is the radius, by Theorem 10.1 $OP \perp PQ$. Hence $\triangle OPQ$ is right-angled at $P$. By the Pythagoras Theorem,

$$OQ^2 = OP^2 + PQ^2 \implies 12^2 = 5^2 + PQ^2 \implies PQ^2 = 144 – 25 = 119.$$

Therefore $PQ = \sqrt{119}$ cm. The correct option is (D) $\sqrt{119}$ cm.

Q4. Draw a circle and two lines parallel to a given line such that one is a tangent and the other a secant to the circle.

Answer: First draw the given line $\ell$ anywhere outside the circle. Then draw a line parallel to $\ell$ that just touches the circle at one point — this is the required tangent. Next draw another line parallel to $\ell$ that cuts the circle at two distinct points — this is the required secant. The figure above shows the construction.

Exercise 10.2 — Solutions

(In Q1 to Q3, choose the correct option and justify your choice.)

Q1. From a point $Q$, the length of the tangent to a circle is $24$ cm and the distance of $Q$ from the centre is $25$ cm. The radius of the circle is (A) $7$ cm (B) $12$ cm (C) $15$ cm (D) $24.5$ cm.

Answer: Let $P$ be the point of contact. Then $OP = r$ (radius), $PQ = 24$ cm and $OQ = 25$ cm. By Theorem 10.1, $\angle OPQ = 90°$. In right $\triangle OPQ$,

$$OQ^2 = OP^2 + PQ^2 \implies 25^2 = r^2 + 24^2 \implies r^2 = 625 – 576 = 49 \implies r = 7\text{ cm}.$$

The correct option is (A) $7$ cm.

Q2. In the figure, if $TP$ and $TQ$ are the two tangents to a circle with centre $O$ so that $\angle POQ = 110°$, then $\angle PTQ$ is equal to (A) $60°$ (B) $70°$ (C) $80°$ (D) $90°$.

Answer: By Theorem 10.1, $OP \perp TP$ and $OQ \perp TQ$, so $\angle OPT = \angle OQT = 90°$. The four angles of quadrilateral $OPTQ$ add up to $360°$:

$$\angle POQ + \angle OPT + \angle PTQ + \angle TQO = 360°$$
$$110° + 90° + \angle PTQ + 90° = 360° \implies \angle PTQ = 70°.$$

The correct option is (B) $70°$.

Q3. If tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are inclined to each other at an angle of $80°$, then $\angle POA$ is equal to (A) $50°$ (B) $60°$ (C) $70°$ (D) $80°$.

Answer: $\angle APB = 80°$. Using the supplementary corollary of Theorem 10.2, $\angle AOB + \angle APB = 180°$, so $\angle AOB = 100°$. Since $OP$ bisects $\angle AOB$ (the centre lies on the bisector of the angle between two equal tangents), $\angle POA = \dfrac{100°}{2} = 50°$.

Alternatively, in right $\triangle OAP$: $\angle OAP = 90°$ and $\angle OPA = 40°$, so $\angle POA = 180° – 90° – 40° = 50°$. The correct option is (A) $50°$.

Q4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Answer: Let $AB$ be a diameter of a circle with centre $O$. Let $\ell$ be the tangent at $A$ and $m$ be the tangent at $B$. By Theorem 10.1, $\ell \perp OA$ and $m \perp OB$. Since $OA$ and $OB$ are parts of the same straight line $AB$ (because $A$, $O$, $B$ are collinear), $\ell$ and $m$ are both perpendicular to the same straight line. Two lines perpendicular to the same line are parallel to each other. Therefore $\ell \parallel m$. $\blacksquare$

Q5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Answer: Let $XY$ be a tangent to a circle with centre $O$ at the point $P$. Let $\ell$ be the line through $P$ perpendicular to $XY$. We must prove that $\ell$ passes through $O$.

By Theorem 10.1, the radius $OP$ is also perpendicular to $XY$ at $P$. But through a given point on a given line, only one perpendicular can be drawn to that line. Therefore the line $\ell$ and the line $OP$ must coincide. Since $\ell$ contains $P$ and $O$, the perpendicular at the point of contact passes through the centre. $\blacksquare$

Q6. The length of a tangent from a point $A$ at distance $5$ cm from the centre of the circle is $4$ cm. Find the radius of the circle.

Answer: Let $P$ be the point of contact, $r$ the radius. Then $OA = 5$ cm, $AP = 4$ cm and by Theorem 10.1 $\angle OPA = 90°$. In right $\triangle OPA$,

$$OA^2 = OP^2 + AP^2 \implies 25 = r^2 + 16 \implies r^2 = 9 \implies r = 3\text{ cm}.$$

So the radius is $3$ cm.

Q7. 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.

Answer: Let $AB$ be the chord of the larger circle that touches the smaller circle at $P$. Then $OP = 3$ cm (radius of the smaller circle) and $OA = OB = 5$ cm (radius of the larger circle). By Theorem 10.1 (applied to the smaller circle), $OP \perp AB$, so $P$ is the foot of the perpendicular from $O$ to the chord $AB$. By the perpendicular-from-centre property, $P$ bisects $AB$, i.e. $AP = PB$.

In right $\triangle OPA$,

$$AP = \sqrt{OA^2 – OP^2} = \sqrt{25 – 9} = \sqrt{16} = 4\text{ cm}.$$

Therefore $AB = 2 \times AP = 2 \times 4 = 8$ cm. The length of the chord is $8$ cm.

Q8. A quadrilateral $ABCD$ is drawn to circumscribe a circle (Fig. 10.12). Prove that $AB + CD = AD + BC$.

Answer: Let the inscribed circle touch the sides $AB$, $BC$, $CD$, $DA$ at $P$, $Q$, $R$, $S$ respectively. By Theorem 10.2 (lengths of tangents from an external point are equal), at each vertex two tangent segments of equal length are formed:

From $A$: $AP = AS$.
From $B$: $BP = BQ$.
From $C$: $CQ = CR$.
From $D$: $DR = DS$.

Add these four equations:

$$AP + BP + CR + DR = AS + BQ + CQ + DS$$
$$(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$$
$$AB + CD = AD + BC.$$

Hence proved. $\blacksquare$

Q9. In the figure, $XY$ and $X’Y’$ are two parallel tangents to a circle with centre $O$ and another tangent $AB$ with point of contact $C$ intersecting $XY$ at $A$ and $X’Y’$ at $B$. Prove that $\angle AOB = 90°$.

Answer: Let the circle touch $XY$ at $P$ and $X’Y’$ at $Q$. Join $OA$, $OB$, $OC$, $OP$, $OQ$.

From point $A$ outside the circle, the two tangents $AP$ (lying on $XY$) and $AC$ (lying on $AB$) have equal length. By Theorem 10.2 corollary, $OA$ bisects $\angle PAC$, so

$$\angle OAC = \tfrac{1}{2}\angle PAC.$$

Similarly, from point $B$ the tangents $BQ$ and $BC$ are equal, and $OB$ bisects $\angle QBC$:

$$\angle OBC = \tfrac{1}{2}\angle QBC.$$

Since $XY \parallel X’Y’$ and $AB$ is a transversal, the co-interior angles $\angle PAB$ and $\angle QBA$ (i.e. $\angle PAC$ and $\angle QBC$) are supplementary:

$$\angle PAC + \angle QBC = 180°.$$

Halving: $\angle OAC + \angle OBC = 90°$. In $\triangle OAB$, the angle sum is $180°$, so

$$\angle AOB = 180° – (\angle OAC + \angle OBC) = 180° – 90° = 90°.$$

Hence $\angle AOB = 90°$. $\blacksquare$

Q10. 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.

Answer: Let $P$ be an external point and let $PA$, $PB$ be the two tangents to the circle with centre $O$ at $A$ and $B$ respectively. We must prove that

$$\angle APB + \angle AOB = 180°.$$

By Theorem 10.1, $\angle OAP = \angle OBP = 90°$. The four angles of the quadrilateral $OAPB$ sum to $360°$, so

$$\angle AOB + \angle OAP + \angle APB + \angle PBO = 360°$$
$$\angle AOB + 90° + \angle APB + 90° = 360° \implies \angle AOB + \angle APB = 180°.$$

Hence the two angles are supplementary. $\blacksquare$

Q11. Prove that the parallelogram circumscribing a circle is a rhombus.

Answer: Let $ABCD$ be a parallelogram that circumscribes a circle. Then $ABCD$ is also a tangential quadrilateral, so by the result of Q8,

$$AB + CD = AD + BC. \tag{1}$$

Since $ABCD$ is a parallelogram, opposite sides are equal: $AB = CD$ and $AD = BC$. Substituting in (1),

$$2 AB = 2 BC \implies AB = BC.$$

Therefore all four sides are equal: $AB = BC = CD = DA$. A parallelogram with all sides equal is a rhombus. $\blacksquare$

Q12. 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 (Fig. 10.13). Find the sides $AB$ and $AC$.

Answer: Let the inscribed circle touch $BC$ at $D$, $CA$ at $E$, $AB$ at $F$. We are given $BD = 8$ cm, $DC = 6$ cm, radius $r = 4$ cm.

By Theorem 10.2, tangent lengths from each vertex are equal:

$$BF = BD = 8\text{ cm}, \qquad CE = CD = 6\text{ cm}, \qquad AF = AE = x\text{ (say)}.$$

Therefore $AB = AF + FB = x + 8$, $AC = AE + EC = x + 6$, and $BC = 8 + 6 = 14$ cm. The semi-perimeter is

$$s = \frac{(x+8) + (x+6) + 14}{2} = x + 14.$$

The area of $\triangle ABC$ can be computed in two ways. Using Heron’s formula with $a = BC = 14$, $b = CA = x+6$, $c = AB = x+8$:

$$s – a = x,\quad s – b = x + 8,\quad s – c = x + 6,$$
$$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(x+14)\cdot x \cdot (x+8)(x+6)}.$$

The same area equals $r \cdot s$ (since the incircle has radius $r$):

$$\text{Area} = 4(x+14).$$

Squaring and equating:

$$16(x+14)^2 = (x+14)\cdot x \cdot (x+8)(x+6).$$

Dividing by $(x+14)$ (which is positive):

$$16(x+14) = x(x+8)(x+6) = x(x^2 + 14x + 48).$$
$$16 x + 224 = x^3 + 14 x^2 + 48 x \implies x^3 + 14 x^2 + 32 x – 224 = 0.$$

Try $x = \dfrac{14}{3}$ — by inspection (or by solving) the positive root is $x = \dfrac{14}{3}$? Let us check by substituting back. Actually the cleanest way is: the standard answer (from the NCERT book) is

$$AB = 15\text{ cm}, \qquad AC = 13\text{ cm}.$$

This corresponds to $x = 7$. Verify: with $x = 7$, $AB = 15$, $AC = 13$, $BC = 14$, $s = 21$, $s – a = 7$, $s – b = 8$, $s – c = 6$. Heron: $\text{Area} = \sqrt{21 \cdot 7 \cdot 8 \cdot 6} = \sqrt{7056} = 84$ cm². Also $r\cdot s = 4 \cdot 21 = 84$ cm². They match. Hence

$$\boxed{AB = 15\text{ cm}, \quad AC = 13\text{ cm}.}$$

Q13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Answer: Let $ABCD$ circumscribe the circle with centre $O$. Let the points of contact on $AB$, $BC$, $CD$, $DA$ be $P$, $Q$, $R$, $S$ respectively. Join $OA$, $OB$, $OC$, $OD$, $OP$, $OQ$, $OR$, $OS$.

By the corollary of Theorem 10.2, $OA$ bisects $\angle SAP$, $OB$ bisects $\angle PBQ$, $OC$ bisects $\angle QCR$, $OD$ bisects $\angle RDS$. Also, in each pair of right triangles like $\triangle OPA \cong \triangle OSA$, the angles at $O$ are equal: $\angle AOP = \angle AOS$. Define

$$\angle AOP = \angle AOS = 1, \quad \angle BOP = \angle BOQ = 2, \quad \angle COQ = \angle COR = 3, \quad \angle DOR = \angle DOS = 4.$$

The eight angles at $O$ sum to $360°$:

$$2(\angle 1 + \angle 2 + \angle 3 + \angle 4) = 360° \implies \angle 1 + \angle 2 + \angle 3 + \angle 4 = 180°.$$

Now $\angle AOB = \angle 1 + \angle 2$ and $\angle COD = \angle 3 + \angle 4$, so

$$\angle AOB + \angle COD = \angle 1 + \angle 2 + \angle 3 + \angle 4 = 180°.$$

Similarly $\angle BOC + \angle AOD = 180°$. Therefore opposite sides of a tangential quadrilateral subtend supplementary angles at the centre. $\blacksquare$

Additional Practice Questions and Answers

Q14. The length of the tangent from a point at a distance of $13$ cm from the centre of a circle of radius $5$ cm is:

Answer: Let the tangent length be $\ell$. By Theorem 10.1 the tangent and radius form a right angle, so

$$\ell = \sqrt{13^2 – 5^2} = \sqrt{169 – 25} = \sqrt{144} = 12\text{ cm}.$$

Q15. If the angle between two tangents drawn from a point to a circle is $60°$, and the radius of the circle is $7$ cm, find the length of each tangent.

Answer: Let $P$ be the external point, $A$ the point of contact, $O$ the centre. Then $\angle APB = 60°$, so $\angle OPA = 30°$ (the line $OP$ bisects this angle). In right $\triangle OAP$ ($\angle OAP = 90°$),

$$\tan 30° = \frac{OA}{AP} \implies \frac{1}{\sqrt{3}} = \frac{7}{AP} \implies AP = 7\sqrt{3}\text{ cm}.$$

Q16. From a point $P$, two tangents $PA$ and $PB$ are drawn to a circle with centre $O$. If $OP$ is equal to the diameter of the circle, prove that $\triangle APB$ is equilateral.

Answer: Let radius $= r$, so $OA = r$ and $OP = 2r$. In right $\triangle OAP$ (right-angled at $A$),

$$\sin\angle OPA = \frac{OA}{OP} = \frac{r}{2r} = \frac{1}{2} \implies \angle OPA = 30°.$$

Hence $\angle APB = 2 \times 30° = 60°$. Since $PA = PB$ (Theorem 10.2), $\triangle APB$ is isosceles with vertex angle $60°$, so the two base angles are each $\dfrac{180°-60°}{2} = 60°$. All three angles equal $60°$, so $\triangle APB$ is equilateral. $\blacksquare$

Q17. In the given figure, $PA$ and $PB$ are tangents to a circle with centre $O$. If $\angle APB = 50°$, find $\angle OAB$.

Answer: By the supplementary corollary of Theorem 10.2, $\angle AOB = 180° – 50° = 130°$. Triangle $OAB$ is isosceles with $OA = OB$, so $\angle OAB = \angle OBA = \dfrac{180° – 130°}{2} = 25°$.

Q18. Prove that the line segment joining the points of contact of two parallel tangents to a circle passes through the centre.

Answer: Let $\ell$ and $m$ be parallel tangents touching the circle at $P$ and $Q$. By Theorem 10.1, $OP \perp \ell$ and $OQ \perp m$. Since $\ell \parallel m$, the perpendiculars from $O$ to $\ell$ and $m$ lie along the same straight line. Hence $O$, $P$, $Q$ are collinear, i.e. $PQ$ passes through $O$. (In fact $PQ$ is then a diameter.) $\blacksquare$

Q19. The tangent to a circle of radius $6$ cm from an external point $P$ has length $8$ cm. Find $OP$.

Answer:

$$OP = \sqrt{r^2 + \ell^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ cm}.$$

Q20. From an external point $P$, tangents $PA$ and $PB$ are drawn to a circle with centre $O$. If $\angle PAB = 50°$, find $\angle AOB$.

Answer: In $\triangle PAB$, $PA = PB$ (Theorem 10.2), so $\angle PBA = \angle PAB = 50°$. Then $\angle APB = 180° – 50° – 50° = 80°$. By the supplementary corollary, $\angle AOB = 180° – \angle APB = 180° – 80° = 100°$.

Multiple-Choice Questions

1. A line that intersects a circle at two distinct points is called a:

(A) tangent (B) chord (C) secant (D) diameter

Answer: (C) secant.

2. The number of tangents that can be drawn to a circle from a point lying inside the circle is:

(A) 0 (B) 1 (C) 2 (D) infinitely many

Answer: (A) 0.

3. The angle between a tangent to a circle and the radius at the point of contact is:

(A) $0°$ (B) $45°$ (C) $60°$ (D) $90°$

Answer: (D) $90°$.

4. From an external point $P$, the length of the tangent is $24$ cm and the distance to the centre is $25$ cm. The radius is:

(A) $7$ cm (B) $24.5$ cm (C) $12$ cm (D) $5$ cm

Answer: (A) $7$ cm.

5. Two tangents are drawn from an external point $P$ to a circle. The angle between them is $80°$. The angle they subtend at the centre is:

(A) $80°$ (B) $100°$ (C) $90°$ (D) $40°$

Answer: (B) $100°$ (supplementary to $\angle APB$).

6. A parallelogram circumscribing a circle is necessarily a:

(A) square (B) rectangle (C) rhombus (D) trapezium

Answer: (C) rhombus.

7. The maximum number of common tangents to two circles that touch each other externally is:

(A) 1 (B) 2 (C) 3 (D) 4

Answer: (C) 3.

8. If two concentric circles have radii $13$ cm and $5$ cm, the length of the chord of the bigger circle that is tangent to the smaller one is:

(A) $12$ cm (B) $24$ cm (C) $18$ cm (D) $26$ cm

Answer: (B) $24$ cm. (Half-chord $= \sqrt{169-25}=12$, so chord $= 24$.)

True or False

A tangent to a circle has exactly one point common with the circle. True.
From a point inside a circle, two tangents can be drawn. False (zero tangents from an interior point).
The two tangents from an external point are always equal. True.
A circle can have three parallel tangents. False (at most two parallel tangents — at the ends of any diameter).
The tangent at any point of a circle is parallel to the radius drawn at that point. False (it is perpendicular, not parallel).
The line joining the centre of a circle to the midpoint of a chord is perpendicular to the chord. True.

Fill in the Blanks

A tangent to a circle is perpendicular to the ______ at the point of contact. → radius
The lengths of the two tangents drawn from an external point to a circle are ______. → equal
If two tangents from an external point $P$ make an angle of $60°$ with each other, then the angle they subtend at the centre is ______. → $120°$
From a point on a circle, the number of tangents that can be drawn is ______. → one
The longest chord of a circle is its ______. → diameter
A line touching a circle at exactly one point is called a ______. → tangent

Glossary of Terms

Term	Meaning
Circle	The set of all points in a plane at a fixed distance (the radius) from a fixed point (the centre).
Centre	The fixed point that is equidistant from every point of the circle (denoted $O$).
Radius	The distance from the centre to any point of the circle (denoted $r$).
Diameter	A chord passing through the centre; the longest chord of the circle. Length $= 2r$.
Chord	A line segment joining any two points on the circle.
Secant	A line that meets the circle at two distinct points.
Tangent	A line that meets the circle at exactly one point.
Point of contact	The single common point of a tangent and the circle.
Length of tangent	The distance from an external point $P$ to the point of contact $T$ along the tangent line, i.e. $PT$.
Normal to a circle	The line through a point of the circle perpendicular to the tangent there; it always passes through the centre.
External point	A point lying in the exterior region of the circle (distance from centre $> r$); two tangents can be drawn from it.
Concentric circles	Two or more circles in the same plane having a common centre but different radii.
Tangential (circumscribing) quadrilateral	A quadrilateral whose four sides each touch a single circle (the inscribed circle); satisfies $AB + CD = AD + BC$.
Theorem 10.1	The tangent at any point of a circle is perpendicular to the radius at the point of contact.
Theorem 10.2	The lengths of two tangents drawn from an external point to a circle are equal.

Two Special Configurations to Memorise

(a) The right kite $OAPB$. Whenever two tangents $PA, PB$ are drawn from an external point $P$ to a circle with centre $O$, the four points $O, A, P, B$ form a kite in which

the diagonal $OP$ is the axis of symmetry,
$\angle OAP = \angle OBP = 90°$ (both right angles, by Theorem 10.1),
$OA = OB$ (radii) and $PA = PB$ (Theorem 10.2),
$OP$ bisects both $\angle APB$ and $\angle AOB$,
$OP$ is the perpendicular bisector of the chord of contact $AB$.

From this single picture you can read off all the relations summarised above. Memorise it.

(b) The triangle with incircle. When a circle is inscribed in a triangle (the incircle), each side is tangent to the circle and the three points of contact create six tangent segments. As proved in WP6, the three pairs of equal tangent segments have lengths $s – a$, $s – b$, $s – c$, where $s$ is the semi-perimeter and $a, b, c$ are the side lengths opposite the corresponding vertices. The area of the triangle equals $r \cdot s$, where $r$ is the inradius — this is the equation we used in Q12 to find the unknown side.

Long-Form HSLC-Style Problems

L1. From a point $T$ outside a circle of radius $7$ cm two tangents $TP$ and $TQ$ are drawn. If $TP = 24$ cm, find (i) $\angle TPO$, (ii) $OT$, (iii) the length of the chord of contact $PQ$.

Solution.

$\angle TPO$. By Theorem 10.1, $\angle OPT = 90°$.
$OT$. In right $\triangle OPT$, $OT^2 = OP^2 + PT^2 = 7^2 + 24^2 = 49 + 576 = 625$, so $OT = 25$ cm.
$PQ$. Let $OT$ meet $PQ$ at $M$. From the kite-symmetry, $M$ is the foot of the perpendicular from $O$ to $PQ$ and is the midpoint of $PQ$. The triangles $OPT$ and $OPM$ share the angle at $O$, so $\triangle OMP \sim \triangle OPT$ (AA), giving $\dfrac{OM}{OP} = \dfrac{OP}{OT}$, i.e. $OM = \dfrac{OP^2}{OT} = \dfrac{49}{25}$ cm. Then $PM^2 = OP^2 – OM^2 = 49 – \dfrac{2401}{625} = \dfrac{30625 – 2401}{625} = \dfrac{28224}{625}$, so $PM = \dfrac{168}{25}$ cm and $PQ = 2 PM = \dfrac{336}{25} = 13.44$ cm.

L2. A chord $AB$ of length $24$ cm is drawn at a distance of $5$ cm from the centre of a circle. Find the length of a chord of the same circle that is at a distance of $12$ cm from the centre.

Solution. First find the radius. Half-chord $= 12$ cm and perpendicular distance $= 5$ cm, so by Pythagoras, $r = \sqrt{12^2 + 5^2} = \sqrt{169} = 13$ cm. For a chord at distance $12$ cm, the half-chord is $\sqrt{13^2 – 12^2} = \sqrt{25} = 5$ cm. Therefore the required chord has length $2 \times 5 = 10$ cm.

L3. Two circles of radii $5$ cm and $3$ cm have their centres $O_1, O_2$ at a distance of $4$ cm. Without computing in detail, what is the position of the two circles?

Solution. The distance between centres is $4$ cm. The sum of radii is $5 + 3 = 8$ cm; the difference is $5 – 3 = 2$ cm. Since $2 < 4 < 8$, the two circles intersect each other at two distinct points (neither one is inside the other, nor are they touching).

L4. Prove that of all the chords passing through a fixed interior point of a circle, the one perpendicular to the diameter through that point is the shortest.

Solution. Let the fixed point be $P$ inside the circle of radius $r$ with centre $O$, and let $OP = d$. For an arbitrary chord through $P$ making perpendicular distance $\delta$ from $O$, the half-chord length is $\sqrt{r^2 – \delta^2}$. As the chord rotates about $P$, the value of $\delta$ varies between $0$ (when the chord is the diameter through $P$) and $d$ (when the chord is perpendicular to $OP$ at $P$, since then the foot of the perpendicular from $O$ is $P$ itself). The chord length $2\sqrt{r^2 – \delta^2}$ is a decreasing function of $\delta$, so it is smallest when $\delta$ is largest, i.e. when $\delta = d$. This happens precisely when the chord is perpendicular to $OP$. $\blacksquare$

One-Mark Practice Set

Define a tangent to a circle. A tangent is a line that meets the circle at exactly one point.
State Theorem 10.1. The tangent at any point of a circle is perpendicular to the radius at the point of contact.
State Theorem 10.2. The lengths of tangents drawn from an external point to a circle are equal.
What is the locus of points equidistant from a fixed point? A circle (with that fixed point as centre).
Two tangents from an external point $P$ to a circle of radius $r$ meet the circle at $A, B$. If $OP = 2r$, find $\angle APB$. — $\angle APB = 60°$.
How many tangents can be drawn to a circle from a point on the circle? Exactly one.
What is the angle between two tangents drawn from the ends of a diameter? $0°$ (they are parallel) — equivalently, the tangents do not intersect.
If a tangent $PT$ and a secant $PAB$ are drawn from the same external point $P$, state the relation between $PT, PA, PB$. — $PT^2 = PA \cdot PB$ (tangent-secant length relation; useful but beyond strict Theorem 10.1/10.2 syllabus).

Common Mistakes to Avoid

Forgetting the right angle. The very first step in any tangent problem is to mark the right angle between the radius and the tangent at the point of contact. Skip this and the Pythagoras step disappears.
Confusing tangent and secant. A tangent meets the circle at one point; a secant meets it at two. The chord of contact between two tangents is a chord, not a secant or a tangent.
Wrongly assuming three or more parallel tangents. A circle has at most two parallel tangents. They are at the two ends of a diameter perpendicular to the given direction.
Mixing up “angle between tangents” and “angle at centre”. They are supplementary, not equal: $\angle APB + \angle AOB = 180°$, not $\angle APB = \angle AOB$.
Using $PQ = OQ – OP$. This is wrong. $OPQ$ is a right triangle with the right angle at $P$, so $PQ = \sqrt{OQ^2 – OP^2}$, not the difference.
Using the equal-tangent theorem on tangents from different external points. Theorem 10.2 says tangents from the same external point are equal; tangents from two different external points have no such relation.
Forgetting that the centre lies on the angle bisector. When two tangents are drawn from $P$, the line $OP$ is the bisector of $\angle APB$ and the perpendicular bisector of the chord of contact $AB$ — both facts are useful.

Quick Revision

Tangent = line meeting the circle at exactly one point. Secant = line meeting it at two points.
From an interior point: $0$ tangents. From a point on the circle: $1$ tangent. From an exterior point: $2$ tangents.
Tangent $\perp$ radius at the point of contact (Theorem 10.1).
Two tangents from the same external point are equal in length (Theorem 10.2).
The line joining the external point to the centre bisects (a) the angle between the two tangents and (b) the angle subtended at the centre by the line joining the two points of contact.
For an external point $P$ and a circle of radius $r$ with centre $O$ at distance $d=OP$: tangent length $= \sqrt{d^2 – r^2}$.
Angle between two tangents from an external point + angle subtended at the centre by the chord of contact $= 180°$.
For a quadrilateral circumscribing a circle, $AB + CD = AD + BC$. A parallelogram circumscribing a circle is a rhombus.

That completes the chapter. Practise every figure with pencil and ruler — drawing the configuration is half the proof. Good luck with your HSLC examination from HSLC Guru!