## Chapter 10 Circles

### 10.1 Introduction

You have studied in Class IX that a circle is a collection of all points in a plane which are at a constant distance (radius) from a fixed point (centre). You have also studied various terms related to a circle like chord, segment, sector, arc etc. Let us now examine the different situations that can arise when a circle and a line are given in a plane.

So, let us consider a circle and a line PQ. There can be three possibilities given in Fig. 10.1 below:

Fig. 10.1

In Fig. 10.1 (i), the line PQ and the circle have no common point. In this case, PQ is called a non-intersecting line with respect to the circle. In Fig. 10.1 (ii), there are two common points $\mathrm{A}$ and $\mathrm{B}$ that the line $\mathrm{PQ}$ and the circle have. In this case, we call the line PQ a secant of the circle. In Fig. 10.1 (iii), there is only one point A which is common to the line PQ and the circle. In this case, the line is called a tangent to the circle.

You might have seen a pulley fitted over a well which is used in taking out water from the well. Look at Fig. 10.2. Here the rope on both sides of the pulley, if considered as a ray, is like a tangent to the circle representing the pulley.

Fig. 10.2

Is there any position of the line with respect to the circle other than the types given above? You can see that there cannot be any other type of position of the line with respect to the circle. In this chapter, we will study about the existence of the tangents to a circle and also study some of their properties.

### 10.2 Tangent to a Circle

In the previous section, you have seen that a tangent[^0] to a circle is a line that intersects the circle at only one point.

To understand the existence of the tangent to a circle at a point, let us perform the following activities:

Activity 1: Take a circular wire and attach a straight wire $A B$ at a point $P$ of the circular wire so that it can rotate about the point $\mathrm{P}$ in a plane. Put the system on a table and gently rotate the wire $\mathrm{AB}$ about the point $\mathrm{P}$ to get different positions of the straight wire [see Fig. 10.3(i)].

In various positions, the wire intersects the circular wire at $\mathrm{P}$ and at another point $\mathrm{Q_1}$ or $\mathrm{Q_2}$ or $\mathrm{Q_3}$, etc. In one position, you will see that it will intersect the circle at the point $\mathrm{P}$ only (see position $\mathrm{A}^{\prime} \mathrm{B}^{\prime}$ of $\mathrm{AB}$ ). This shows that a tangent exists at the point $\mathrm{P}$ of the circle. On rotating further, you can observe that in all other positions of $\mathrm{AB}$, it will intersect the circle at $\mathrm{P}$ and at another point, say $\mathrm{R_1}$ or $\mathrm{R_2}$ or $\mathrm{R_3}$, etc. So, you can observe that there is only one tangent at a point of the circle.

Fig. 10.3 (i)

While doing activity above, you must have observed that as the position $A B$ moves towards the position $\mathrm{A}^{\prime} \mathrm{B}^{\prime}$, the common point, say $\mathrm{Q_1}$, of the line $\mathrm{AB}$ and the circle gradually comes nearer and nearer to the common point $\mathrm{P}$. Ultimately, it coincides with the point $\mathrm{P}$ in the position $\mathrm{A}^{\prime} \mathrm{B}^{\prime}$ of $\mathrm{A}^{\prime \prime} \mathrm{B}^{\prime}$. Again note, what happens if ’ $\mathrm{AB}$ ’ is rotated rightwards about $\mathrm{P}$ ? The common point $\mathrm{R_3}$ gradually comes nearer and nearer to P and ultimately coincides with P. So, what we see is:

The tangent to a circle is a special case of the secant, when the two end points of its corresponding chord coincide.

Activity 2: On a paper, draw a circle and a secant PQ of the circle. Draw various lines parallel to the secant on both sides of it. You will find that after some steps, the length of the chord cut by the lines will gradually decrease, i.e., the two points of intersection of the line and the circle are coming closer and closer [see Fig. 10.3(ii)]. In one case, it becomes zero on one side of the secant and in another case, it becomes zero on the other side of the secant. See the positions $\mathrm{P}^{\prime} \mathrm{Q}^{\prime}$ and $\mathrm{P}^{\prime \prime} \mathrm{Q}^{\prime \prime}$ of the secant in Fig. 10.3 (ii). These are the tangents to the circle parallel to the given secant PQ. This also helps you to see that there cannot be more than two tangents parallel to a given secant.

Fig. 10.3 (ii)

This activity also establishes, what you must have observed, while doing Activity 1, namely, a tangent is the secant when both of the end points of the corresponding chord coincide.

The common point of the tangent and the circle is called the point of contact [the point A in Fig. 10.1 (iii)]and the tangent is said to touch the circle at the common point.

Now look around you. Have you seen a bicycle or a cart moving? Look at its wheels. All the spokes of a wheel are along its radii. Now note the position of the wheel with respect to its movement on the ground. Do you see any tangent anywhere? (See Fig. 10.4). In fact, the wheel moves along a line which is a tangent to the circle representing the wheel. Also, notice that in all positions, the radius through the point of contact with the ground appears to be at right angles to the tangent (see Fig. 10.4). We shall now prove this property of the tangent.

Fig. 10.4

**Theorem 10.1** : The tangent at any point of a circle is perpendicular to the radius through the point of contact.

**Proof** : We are given a circle with centre $\mathrm{O}$ and a tangent $\mathrm{XY}$ to the circle at a point $\mathrm{P}$. We need to prove that $\mathrm{OP}$ is perpendicular to $\mathrm{XY}$.

Take a point $\mathrm{Q}$ on $\mathrm{XY}$ other than $\mathrm{P}$ and join $\mathrm{OQ}$ (see Fig. 10.5).

The point $\mathrm{Q}$ must lie outside the circle. (Why? Note that if Q lies inside the circle, XY will become a secant and not a tangent to the circle). Therefore, OQ is longer than the radius $\mathrm{OP}$ of the circle. That is,

$$ \mathrm{OQ}>\mathrm{OP} . $$

Since this happens for every point on the line $\mathrm{XY}$ except the point $\mathrm{P}$, $\mathrm{OP}$ is the shortest of all the distances of the point $\mathrm{O}$ to the points of XY. So OP is perpendicular to XY. (as shown in Theorem A1.7.)

Fig. 10.5

**Remarks**

**1.** By theorem above, we can also conclude that at any point on a circle there can be one and only one tangent.

**2.** The line containing the radius through the point of contact is also sometimes called the ’normal’ to the circle at the point.

### EXERCISE 10.1

**1.** How many tangents can a circle have?

## Show Answer

**Solution**

A circle can have infinite tangents.

**2.** 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 _______.

## Show Answer

**Solution**

(i) One

(ii) Secant

(iii) Two

(iv) Point of contact

**3.** A tangent $\mathrm{PQ}$ at a point $\mathrm{P}$ of a circle of radius $5 \mathrm{~cm}$ meets a line through the centre $\mathrm{O}$ at a point $Q$ so that $O Q=12 \mathrm{~cm}$. Length $P Q$ is :

(A) $12 \mathrm{~cm}$

(B) $13 \mathrm{~cm}$

(C) $8.5 \mathrm{~cm}$

(D) $\sqrt{119} \mathrm{~cm}$.

## Show Answer

**Solution**

We know that the line drawn from the centre of the circle to the tangent is perpendicular to the tangent.

$\therefore OP \perp PQ$

By applying Pythagoras theorem in $\triangle OPQ$,

$\therefore OP^{2}+PQ^{2}=OQ^{2}$

$5^{2}+PQ^{2}=12^{2}$

$PQ^{2}=144-25$

$PQ=\sqrt{119} cm$.

Hence, the correct answer is (D).

**4.** 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.

## Show Answer

**Solution**

It can be observed that $A B$ and $C D$ are two parallel lines. Line $A B$ is intersecting the circle at exactly two points, $P$ and $Q$. Therefore, line $A B$ is the secant of this circle. Since line $C D$ is intersecting the circle at exactly one point, $R$, line $C D$ is the tangent to the circle.

### 10.3 Number of Tangents from a Point on a Circle

To get an idea of the number of tangents from a point on a circle, let us perform the following activity:

Activity 3: Draw a circle on a paper. Take a point $P$ inside it. Can you draw a tangent to the circle through this point? You will find that all the lines through this point intersect the circle in two points. So, it is not possible to draw any tangent to a circle through a point inside it [see Fig. 10.6 (i)].

Next take a point $\mathrm{P}$ on the circle and draw tangents through this point. You have already observed that there is only one tangent to the circle at such a point [see Fig. 10.6 (ii)].

Finally, take a point $P$ outside the circle and try to draw tangents to the circle from this point. What do you observe? You will find that you can draw exactly two tangents to the circle through this point [see Fig. 10.6 (iii)].

Fig. 10.6

We can summarise these facts as follows:

**Case 1** : There is no tangent to a circle passing through a point lying on the circle.

**Case 2** : There is one and only one tangent to a circle passing through a point lying on the circle.

**Case 3** : There are exactly two tangents to a circle through a point lying outside the circle.

In Fig. 10.6 (iii), $\mathrm{T_1}$ and $\mathrm{T_2}$ are the points of contact of the tangents $\mathrm{PT_1}$ and $\mathrm{PT_2}$ respectively.

The length of the segment of the tangent from the external point $P$ and the point of contact with the circle is called the length of the tangent from the point $\mathrm{P}$ to the circle.

Note that in Fig. 10.6 (iii), $\mathrm{PT_1}$ and $\mathrm{PT_2}$ are the lengths of the tangents from $\mathrm{P}$ to the circle. The lengths $\mathrm{PT_1}$ and $\mathrm{PT_2}$ have a common property. Can you find this? Measure $\mathrm{PT_1}$ and $\mathrm{PT_2}$. Are these equal? In fact, this is always so. Let us give a proof of this fact in the following theorem.

**Theorem 10.2**: The lengths of tangents drawn from an external point to a circle are equal.

**Proof**: We are given a circle with centre $\mathrm{O}$, a point $\mathrm{P}$ lying outside the circle and two tangents PQ, PR on the circle from $\mathrm{P}$ (see Fig. 10.7). We are required to prove that $P Q=P R$.

Fig. 10.7

For this, we join OP, OQ and OR. Then $\angle \mathrm{OQP}$ and $\angle \mathrm{ORP}$ are right angles, because these are angles between the radii and tangents, and according to Theorem 10.1 they are right angles. Now in right triangles OQP and ORP,

OQ $=$ OR $\quad \quad \quad \quad \quad \quad \quad $(Radii of the same circle)

OP $=$ OP $\quad \quad \quad \quad \quad \quad \quad $ (Common)

Therefore, $\Delta \mathrm{OQP} \cong \triangle \mathrm{ORP}\quad \quad \quad \quad \text{(RHS)}$

This gives $P Q=P R\quad \quad \quad \quad \text{(CPCT)}$

**Remarks**

**1.** The theorem can also be proved by using the Pythagoras Theorem as follows:

$ \mathrm{PQ}^{2}=\mathrm{OP}^{2}-\mathrm{OQ}^{2}=\mathrm{OP}^{2}-\mathrm{OR}^{2}=\mathrm{PR}^{2}(\mathrm{As} \mathrm{OQ}=\mathrm{OR}) $

which gives $P Q=P R$.

**2.** Note also that $\angle \mathrm{OPQ}=\angle \mathrm{OPR}$. Therefore, $\mathrm{OP}$ is the angle bisector of $\angle \mathrm{QPR}$, i.e., the centre lies on the bisector of the angle between the two tangents.

Let us take some examples.

**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** : We are given two concentric circles $\mathrm{C_1}$ and $\mathrm{C_2}$ with centre $\mathrm{O}$ and a chord $\mathrm{AB}$ of the larger circle $\mathrm{C_1}$ which touches the smaller circle $\mathrm{C_2}$ at the point $\mathrm{P}$ (see Fig. 10.8). We need to prove that $\mathrm{AP}=\mathrm{BP}$.

Fig. 10.8

Let us join OP. Then, $A B$ is a tangent to $C_{2}$ at $P$ and $\mathrm{OP}$ is its radius. Therefore, by Theorem 10.1,

$$ \mathrm{OP} \perp \mathrm{AB} $$

Now $\mathrm{AB}$ is a chord of the circle $\mathrm{C}_{1}$ and $\mathrm{OP} \perp \mathrm{AB}$. Therefore, $\mathrm{OP}$ is the bisector of the chord $A B$, as the perpendicular from the centre bisects the chord,

i.e., $$ \mathrm{AP}=\mathrm{BP} $$

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

**Solution** : We are given a circle with centre $\mathrm{O}$, an external point $\mathrm{T}$ and two tangents $\mathrm{TP}$ and $\mathrm{TQ}$ to the circle, where $\mathrm{P}, \mathrm{Q}$ are the points of contact (see Fig. 10.9). We need to prove that

Fig. 10.9

$$ \angle \mathrm{PTQ}=2 \angle \mathrm{OPQ} $$

Let $$ \angle \mathrm{PTQ}=\theta $$

Now, by Theorem 10.2, TP = TQ. So, TPQ is an isosceles triangle.

Therefore, $$ \angle \mathrm{TPQ}=\angle \mathrm{TQP}=\dfrac{1}{2}\left(180^{\circ}-\theta\right)=90^{\circ}-\dfrac{1}{2} \theta $$

Also, by Theorem 10.1, $$ \angle \mathrm{OPT}=90^{\circ} $$

So, $$ \begin{aligned} \angle \mathrm{OPQ}=\angle \mathrm{OPT}-\angle \mathrm{TPQ} & =90^{\circ}-\left(90^{\circ}-\dfrac{1}{2} \theta\right) \\ & =\dfrac{1}{2} \theta=\dfrac{1}{2} \angle \mathrm{PTQ} \end{aligned} $$

This gives $$ \angle \mathrm{PTQ}=2 \angle \mathrm{OPQ} $$

**Example 3** : PQ is a chord of length $8 \mathrm{~cm}$ of a circle of radius $5 \mathrm{~cm}$. The tangents at $\mathrm{P}$ and $\mathrm{Q}$ intersect at a point $T$ (see Fig. 10.10). Find the length TP.

**Solution** : Join OT. Let it intersect PQ at the point $\mathrm{R}$. Then $\triangle$ TPQ is isosceles and TO is the angle bisector of $\angle \mathrm{PTQ}$. So, $\mathrm{OT} \perp \mathrm{PQ}$ and therefore, OT bisects $\mathrm{PQ}$ which gives $\mathrm{PR}=\mathrm{RQ}=4 \mathrm{~cm}$.

Fig. 10.10

Also, $\mathrm{OR}=\sqrt{\mathrm{OP}^{2}-\mathrm{PR}^{2}}=\sqrt{5^{2}-4^{2}} \mathrm{~cm}=3 \mathrm{~cm}$.

Now, $\angle \mathrm{TPR}+\angle \mathrm{RPO}=90^{\circ}=\angle \mathrm{TPR}+\angle \mathrm{PTR} \quad$ (Why?)

So, $\quad \angle \mathrm{RPO}=\angle \mathrm{PTR}$

Therefore, right triangle TRP is similar to the right triangle PRO by AA similarity.

This gives

$ \dfrac{\mathrm{TP}}{\mathrm{PO}}=\dfrac{\mathrm{RP}}{\mathrm{RO}} \text {, i.e., } \dfrac{\mathrm{TP}}{5}=\dfrac{4}{3} \text { or } \mathrm{TP}=\dfrac{20}{3} \mathrm{~cm} \text {. } $

**Note :** TP can also be found by using the Pythagoras Theorem, as follows:

Let $$ \begin{array}{rlrl} \mathrm{TP} & =x \text { and } \mathrm{TR}=y . \quad \text { Then } \end{array} $$

$$ \begin{array}{rlrl} x^{2} & =y^{2}+16 & & \quad(\text { Taking right } \Delta \mathrm{PRT}) \tag{1} \end{array} $$

$$ \begin{array}{rlrl} x^{2}+5^{2} & =(y+3)^{2} & & \quad(\text { Taking right } \Delta \mathrm{OPT}) \tag{2} \end{array} $$

Subtracting (1) from (2), we get

Therefore, $$ \begin{aligned} 25 & =6 y-7 \text { or } y=\dfrac{32}{6}=\dfrac{16}{3} \\ x^{2} & =\left(\dfrac{16}{3}\right)^{2}+16=\dfrac{16}{9}(16+9)=\dfrac{16 \times 25}{9} \quad \quad \quad \text{[From (1)]}\\ x & =\dfrac{20}{3} \end{aligned} $$

### EXERCISE 10.2

In Q. 1 to 3, choose the correct option and give justification.

**1.** From a point $\mathrm{Q}$, the length of the tangent to a circle is $24 \mathrm{~cm}$ and the distance of $\mathrm{Q}$ from the centre is $25 \mathrm{~cm}$. The radius of the circle is

(A) $7 \mathrm{~cm}$

(B) $12 \mathrm{~cm}$

(C) $15 \mathrm{~cm}$

(D) $24.5 \mathrm{~cm}$

## Show Answer

**Solution**

Let $O$ be the centre of the circle.

Given that,

$OQ=25 cm$ and $PQ=24 cm$

As the radius is perpendicular to the tangent at the point of contact,

Therefore, $OP \perp PQ$

Applying Pythagoras theorem in $\triangle OPQ$, we obtain

$OP^{2}+PQ^{2}=OQ^{2}$

$OP^{2}+24^{2}=25^{2}$

$OP^{2}=625-576$

$OP^{2}=49$

$OP=7$

Therefore, the radius of the circle is $7 cm$.

Hence, alternative (A) is correct.

**2.** In Fig. 10.11, if $\mathrm{TP}$ and $\mathrm{TQ}$ are the two tangents to a circle with centre $\mathrm{O}$ so that $\angle \mathrm{POQ}=110^{\circ}$, then $\angle \mathrm{PTQ}$ is equal to

Fig. 10.11

(A) $60^{\circ}$

(B) $70^{\circ}$

(C) $80^{\circ}$

(D) $90^{\circ}$

## Show Answer

**Solution**

It is given that TP and TQ are tangents.

Therefore, radius drawn to these tangents will be perpendicular to the tangents.

Thus, $OP \perp TP$ and $OQ \perp TQ$

$\angle OPT=90^{\circ}$

$\angle OQT=90^{\circ}$

In quadrilateral POQT,

Sum of all interior angles $=360^{\circ}$

$\angle OPT+\angle POQ+\angle OQT+\angle PTQ=360^{\circ}$

$\Rightarrow 90^{\circ}+110^{\circ}+90^{\circ}+\angle PTQ=360^{\circ}$

$\Rightarrow \angle PTQ=70^{\circ}$

Hence, alternative (B) is correct.

**3.** If tangents $\mathrm{PA}$ and $\mathrm{PB}$ from a point $\mathrm{P}$ to a circle with centre $\mathrm{O}$ are inclined to each other at angle of $80^{\circ}$, then $\angle \mathrm{POA}$ is equal to

(A) $50^{\circ}$

(B) $60^{\circ}$

(C) $70^{\circ}$

(D) $80^{\circ}$

## Show Answer

**Solution**

It is given that PA and PB are tangents.

Therefore, the radius drawn to these tangents will be perpendicular to the tangents.

Thus, $OA \perp PA$ and $OB \perp PB$

$\angle OBP=90^{\circ}$

$\angle OAP=90^{\circ}$

In AOBP,

Sum of all interior angles $=360^{\circ}$

$ \begin{aligned} & \angle OAP+\angle APB+\angle PBO+\angle BOA=360^{\circ} \\ & 90^{\circ}+80^{\circ}+90^{\circ}+\angle BOA=360^{\circ} \\ & \angle BOA=100^{\circ} \end{aligned} $

In $\triangle OPB$ and $\triangle OPA$,

$A P=B P($ Tangents from a point $)$

$OA=OB$ (Radii of the circle)

$OP=OP$ (Common side)

Therefore, $\triangle OPB \cong \triangle OPA$ (SSS congruence criterion)

$A \rightarrow B, P \rightarrow P, O \rightarrow O$

And thus, $\angle POB=\angle POA$

$\angle POA=\dfrac{1}{2} \angle AOB=\dfrac{100^{\circ}}{2}=50^{\circ}$

Hence, alternative (A) is correct.

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

## Show Answer

**Solution**

Let $A B$ be a diameter of the circle. Two tangents $P Q$ and $R S$ are drawn at points $A$ and $B$ respectively. Radius drawn to these tangents will be perpendicular to the tangents.

Thus, $OA \perp RS$ and $OB \perp PQ$

$\angle OAR=90^{\circ}$

$\angle OAS=90^{\circ}$

$\angle OBP=90^{\circ}$

$\angle OBQ=90^{\circ}$

It can be observed that $\angle OAR=\angle OBQ$ (Alternate interior angles)

$\angle OAS=\angle OBP$ (Alternate interior angles)

Since alternate interior angles are equal, lines $P Q$ and $R S$ will be parallel.

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

## Show Answer

**Solution**

Let us consider a circle with centre $O$. Let $AB$ be a tangent which touches the circle at $P$.

We have to prove that the line perpendicular to $A B$ at $P$ passes through centre $O$. We shall prove this by contradiction method.

Let us assume that the perpendicular to $A B$ at $P$ does not pass through centre $O$. Let it pass through another point $O^{\prime}$. Join OP and O’P.

As perpendicular to $A B$ at $P$ passes through $O^{\prime}$, therefore,

$\angle O^{\prime} PB=90^{\circ}$

$O$ is the centre of the circle and $P$ is the point of contact. We know the line joining the centre and the point of contact to the tangent of the circle are perpendicular to each other.

$\therefore \angle OPB=90^{\circ}$..

Comparing equations (1) and (2), we obtain

$\angle O ’ PB=\angle OPB$..

From the figure, it can be observed that,

$\angle O$ ‘PB $<\angle OPB$…

Therefore, $\angle O O^{\prime} PB=\angle OPB$ is not possible. It is only possible, when the line O’P coincides with OP.

Therefore, the perpendicular to $A B$ through $P$ passes through centre $O$.

**6.** The length of a tangent from a point $A$ at distance $5 \mathrm{~cm}$ from the centre of the circle is 4 $\mathrm{cm}$. Find the radius of the circle.

## Show Answer

**Solution**

Let us consider a circle centered at point $O$.

$A B$ is a tangent drawn on this circle from point $A$.

Given that,

$O A=5 cm$ and $A B=4 cm$

In $\triangle ABO$,

$OB \perp AB$ (Radius $\perp$ tangent at the point of contact)

Applying Pythagoras theorem in $\triangle A B O$, we obtain

$AB^{2}+BO^{2}=OA^{2}$

$4^{2}+BO^{2}=5^{2}$

$16+BO^{2}=25$

$BO^{2}=9$

$BO=3$

Hence, the radius of the circle is $3 cm$.

**7.** Two concentric circles are of radii $5 \mathrm{~cm}$ and $3 \mathrm{~cm}$. Find the length of the chord of the larger circle which touches the smaller circle.

## Show Answer

**Solution**

Let the two concentric circles be centered at point $O$. And let $PQ$ be the chord of the larger circle which touches the smaller circle at point $A$. Therefore, $PQ$ is tangent to the smaller circle.

$OA \perp PQ(As OA$ is the radius of the circle)

Applying Pythagoras theorem in $\triangle OAP$, we obtain

$OA^{2}+AP^{2}=OP^{2}$

$3^{2}+AP^{2}=5^{2}$

$9+AP^{2}=25$

$AP^{2}=16$

$AP=4$

In $\triangle OPQ$,

Since $OA \perp PQ$,

$AP=AQ$ (Perpendicular from the center of the circle bisects the chord)

$\therefore PQ=2 AP=2 \times 4=8$

Therefore, the length of the chord of the larger circle is $8 cm$.

**8.** A quadrilateral $A B C D$ is drawn to circumscribe a circle (see Fig. 10.12). Prove that

$$ \mathrm{AB}+\mathrm{CD}=\mathrm{AD}+\mathrm{BC} $$

Fig. 10.12

## Show Answer

**Solution**

It can be observed that

DR $=$ DS (Tangents on the circle from point D) $CR=CQ$ (Tangents on the circle from point C) …

$B P=B Q($ Tangents on the circle from point $B)$

AP $=$ AS (Tangents on the circle from point A)

Adding all these equations, we obtain

$DR+CR+BP+AP=DS+CQ+BQ+AS$

$(D R+C R)+(B P+A P)=(D S+A S)+(C Q+B Q)$

$C D+A B=A D+B C$

**9.** In Fig. 10.13, $\mathrm{XY}$ and $\mathrm{X}^{\prime} \mathrm{Y}^{\prime}$ are two parallel tangents to a circle with centre $\mathrm{O}$ and another tangent $A B$ with point of contact $C$ intersecting $X Y$ at $A$ and $X^{\prime} Y^{\prime}$ at $B$. Prove that $\angle \mathrm{AOB}=90^{\circ}$.

Fig. 10.13

## Show Answer

**Solution**

Let us join point $O$ to $C$.

In $\triangle OPA$ and $\triangle OCA$,

$OP=OC$ (Radii of the same circle)

$A P=A C($ Tangents from point $A)$

$AO=AO$ (Common side)

$\triangle OPA \cong \triangle OCA$ (SSS congruence criterion)

Therefore, $P \rightarrow C, A \rightarrow A, O \rightarrow O$

$\angle POA=\angle COA \ldots$ (i)

Similarly, $\triangle OQB \cong \triangle OCB$ $\angle QOB=\angle COB \ldots$

Since POQ is a diameter of the circle, it is a straight line.

Therefore, $\angle POA+\angle COA+\angle COB+\angle QOB=180^{\circ}$

From equations (i) and (ii), it can be observed that

$2 \angle COA+2 \angle COB=180^{\circ}$

$\angle COA+\angle COB=90^{\circ}$

$\angle A O B=90^{\circ}$

**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.

## Show Answer

**Solution**

Let us consider a circle centered at point $O$. Let $P$ be an external point from which two tangents $PA$ and $PB$ are drawn to the circle which are touching the circle at point $A$ and $B$ respectively and $A B$ is the line segment, joining point of contacts $A$ and $B$ together such that it subtends $\angle A O B$ at center $O$ of the circle.

It can be observed that

$OA$ (radius) $\perp PA$ (tangent)

Therefore, $\angle OAP=90^{\circ}$

Similarly, OB (radius) $\perp$ PB (tangent)

$\angle OBP=90^{\circ}$

In quadrilateral OAPB,

Sum of all interior angles $=360^{\circ}$

$\angle OAP+\angle APB+\angle PBO+\angle BOA=360^{\circ}$

$90^{\circ}+\angle APB+90^{\circ}+\angle BOA=360^{\circ}$

$\angle APB+\angle BOA=180^{\circ}$

Hence, it can be observed 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.

**11.** Prove that the parallelogram circumscribing a circle is a rhombus.

## Show Answer

**Solution**

Since $A B C D$ is a parallelogram,

$A B=C D$…(1)

$B C=A D \ldots(2)$

It can be observed that

DR $=$ DS (Tangents on the circle from point D)

$CR=CQ$ (Tangents on the circle from point C)

$BP=BQ$ (Tangents on the circle from point $B$ )

AP = AS (Tangents on the circle from point A)

Adding all these equations, we obtain

$DR+CR+BP+AP=DS+CQ+BQ+AS$

$(DR+CR)+(BP+AP)=(DS+AS)+(CQ+BQ)$

$C D+A B=A D+B C$

On putting the values of equations (1) and (2) in this equation, we obtain

$2 AB=2 BC$

$AB=BC$

Comparing equations (1), (2), and (3), we obtain

$AB=BC=CD=DA$

Hence, $A B C D$ is a rhombus.

**12.** A triangle $A B C$ is drawn to circumscribe a circle of radius $4 \mathrm{~cm}$ such that the segments $\mathrm{BD}$ and DC into which $B C$ is divided by the point of contact $\mathrm{D}$ are of lengths $8 \mathrm{~cm}$ and $6 \mathrm{~cm}$ respectively (see Fig. 10.14). Find the sides $A B$ and AC.

Fig. 10.14

## Show Answer

**Solution**

Let the given circle touch the sides $A B$ and $A C$ of the triangle at point $E$ and $F$ respectively and the length of the line segment AF be $x$.

In $\triangle ABC$,

$CF=CD=6 cm$ (Tangents on the circle from point $C$ )

$BE=BD=8 cm$ (Tangents on the circle from point $B$ )

$AE=AF=x($ Tangents on the circle from point $A)$

$AB=AE+EB=x+8$

$BC=BD+DC=8+6=14$

$CA=CF+FA=6+x$

$2 s=A B+B C+C A$

$=x+8+14+6+x$

$=28+2 x$

$s=14+x$

$ \text{ Area of } \begin{aligned} \Delta ABC & =\sqrt{s(s-a)(s-b)(s-c)} \\ & =\sqrt{{14+x}{(14+x)-14}{(14+x)-(6+x)}{(14+x)-(8+x)}} \\ & =\sqrt{(14+x)(x)(8)(6)} \\ & =4 \sqrt{3(14 x+x^{2})} \end{aligned} $

Area of $\triangle OBC=\dfrac{1}{2} \times OD \times BC=\dfrac{1}{2} \times 4 \times 14=28$

Area of $\triangle OCA=\dfrac{1}{2} \times OF \times AC=\dfrac{1}{2} \times 4 \times(6+x)=12+2 x$

Area of $\triangle OAB=\dfrac{1}{2} \times OE \times AB=\dfrac{1}{2} \times 4 \times(8+x)=16+2 x$

Area of $\triangle ABC=$ Area of $\triangle OBC+$ Area of $\triangle OCA+$ Area of $\triangle OAB$

$4 \sqrt{3(14 x+x^{2})}=28+12+2 x+16+2 x$

$\Rightarrow 4 \sqrt{3(14 x+x^{2})}=56+4 x$

$\Rightarrow \sqrt{3(14 x+x^{2})}=14+x$

$\Rightarrow 3(14 x+x^{2})=(14+x)^{2}$

$\Rightarrow 42 x+3 x^{2}=196+x^{2}+28 x$

$\Rightarrow 2 x^{2}+14 x-196=0$

$\Rightarrow x^{2}+7 x-98=0$

$\Rightarrow x^{2}+14 x-7 x-98=0$

$\Rightarrow x(x+14)-7(x+14)=0$

$\Rightarrow(x+14)(x-7)=0$

Either $x+14=0$ or $x-7=0$

Therefore, $x=-14$ and 7

However, $x=-14$ is not possible as the length of the sides will be negative.

Therefore, $x=7$

Hence, $AB=x+8=7+8=15 cm$

$CA=6+x=6+7=13 cm$

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

## Show Answer

**Solution**

Let $A B C D$ be a quadrilateral circumscribing a circle centered at $O$ such that it touches the circle at point $P, Q, R, S$. Let us join the vertices of the quadrilateral $A B C D$ to the center of the circle.

Consider $\triangle OAP$ and $\triangle OAS$,

AP $=$ AS (Tangents from the same point)

$OP=OS$ (Radii of the same circle)

$OA=OA$ (Common side)

$\Delta OAP \cong \triangle OAS$ (SSS congruence criterion)

And thus, $\angle POA=\angle AOS$

$\angle 1=\angle 8$

Similarly,

$\angle 2=\angle 3$

$\angle 4=\angle 5$

$\angle 6=\angle 7$

$\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8=360^{\circ}$

$(\angle 1+\angle 8)+(\angle 2+\angle 3)+(\angle 4+\angle 5)+(\angle 6+\angle 7)=360^{\circ}$

$2 \angle 1+2 \angle 2+2 \angle 5+2 \angle 6=360^{\circ}$

$2(\angle 1+\angle 2)+2(\angle 5+\angle 6)=360^{\circ}$

$(\angle 1+\angle 2)+(\angle 5+\angle 6)=180^{\circ}$

$\angle AOB+\angle COD=180^{\circ}$

Similarly, we can prove that $\angle BOC+\angle DOA=180^{\circ}$

Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

### 10.4 Summary

In this chapter, you have studied the following points :

**1.** The meaning of a tangent to a circle.

**2.** The tangent to a circle is perpendicular to the radius through the point of contact.

**3.** The lengths of the two tangents from an external point to a circle are equal.