Solution

(b) Let $O$ be the centre of two concentric circles $C_1$ and $C_2$, whose radii are $r_1=4 cm$ and $r_2=5 cm$. Now, we draw a chord $A C$ of circle $C_2$, which touches the circle $C_1$ at $B$.

Also, join $O B$, which is perpendicular to $A C$. [Tangent at any point of circle is perpendicular to radius throughly the point of contact]

Now, in right angled $\triangle O B C$, by using Pythagoras theorem,

$ O C^{2}=B C^{2}+B O^{2} $

$\Rightarrow \quad 5^{2}=B C^{2}+4^{2}$

$[\because.$ hypotenuse $.^{2}=(\text{ base })^{2}+(\text{ perpendicular })^{2}]$

$\Rightarrow \quad B C^{2}=25-16=9 \Rightarrow B C=3 cm$

$\therefore$ Length of chord $A C=2 B C=2 \times 3=6 cm$

Solution

(d) We know that, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

i.e.,

$ \begin{aligned} \angle A O B+\angle C O D = 180^{\circ} \\ \angle C O D = 180^{\circ}-\angle A O B \\ & =180^{\circ}-125^{\circ}=55^{\circ} \end{aligned} $

$ \Rightarrow \quad \angle C O D=180^{\circ}-\angle A O B $

Solution

(c) In figure, $A O C$ is a diameter of the circle. We know that, diameter subtends an angle $90^{\circ}$ at the circle.

So,

$ \begin{aligned} \angle A B C = 90^{\circ} \\ \angle A+\angle B+\angle C = 180^{\circ} \end{aligned} $

In $\triangle A C B$,

$\Rightarrow$[since, sum of all angles of a triangle is $180^{\circ}$ ]

$\Rightarrow$ $\angle A+90^{\circ}+50^{\circ}=180^{\circ}$

$\Rightarrow$ $ \angle A+140=180 $

$ \begin{aligned} \angle A=180^{\circ}-140^{\circ} = 40^{\circ} \\ \angle A \text{ or } \angle O A B = 40^{\circ} \end{aligned} $

Now, $A T$ is the tangent to the circle at point $A$. So, $O A$ is perpendicular to $A T$.

$ \begin{aligned} & \therefore \quad \angle O A T=90^{\circ} \quad \text{ [from figure] } \\ & \Rightarrow \quad \angle O A B+\angle B A T=90^{\circ} \\ & \text{ On putting } \angle O A B=40^{\circ} \text{, we get } \\ & \Rightarrow \quad \angle B A T=90^{\circ}-40^{\circ}=50^{\circ} \end{aligned} $

Solution

(a) Firstly, draw a circle of radius $5 cm$ having centre $O . P$ is a point at a distance of $13 cm$ from $O$. A pair of tangents $P Q$ and $P R$ are drawn.

Thus, quadrilateral $P Q O R$ is formed.

$ \begin{matrix} \because & O Q & \perp Q P & \text{ [since, } A P \text{ is a tangent line] } \\ \text{ In right angled } \triangle P Q O, & O P^{2} = O Q^{2}+Q P^{2} \\ \Rightarrow \quad & 13^{2} = 5^{2}+Q P^{2} \\ \Rightarrow \quad & Q P^{2} = 169-25=144 \\ \Rightarrow \quad & Q P = 12 cm \\ & & \\ \text{ Now, } \quad & = \dfrac{1}{2} \times 12 \times 5=30 cm^{2} \\ \therefore \quad & \text{ area of } \triangle O Q P = \dfrac{1}{2} \times Q P \times Q O \\ & & \end{matrix} $

Solution

(d) First, draw a circle of radius $5 cm$ having centre $O$. A tangent $X Y$ is drawn at point $A$.

A chord $C D$ is drawn which is parallel to $X Y$ and at a distance of $8 cm$ from $A$. Now, $\angle O A Y=90^{\circ}$

[Tangent and any point of a circle is perpendicular to the radius through the point of contact] $\Rightarrow$ $ \triangle O A Y+\triangle O E D=180^{\circ} $

Also,

$ \triangle O E D=180^{\circ} $

Now, in right angled $\triangle O E C$,

$ A E=8 cm . \text{ Join } O C $

$\Rightarrow$ $ O C^{2}=O E^{2}+E C^{2} $

$ \begin{matrix} E C^{2} = O C^{2}-O E^{2} & {[\text{ by Pythagoras theorem }]} \\ & =5^{2}-3^{2} & \end{matrix} $

$[\because O C=$ radius $=5 cm, O E=A E-A O=8-5=3 cm]$ $=25-9=16$

$\Rightarrow$ $E C=4 cm$

Hence,

length of chord $C D=2 C E=2 \times 4=8 cm$

[since, perpendicular from centre to the chord bisects the chord]

Solution

(c) Join $O A$.

We know that, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

$\therefore$ $\angle O A T$ $=90^{\circ}$

In $\triangle O A T$, $\cos 30^{\circ}$ $=\dfrac{A T}{O T}$

$\Rightarrow$ $\dfrac{\sqrt{3}}{2}$ $=\dfrac{A T}{4}$

$\Rightarrow$ $A T$ $=2 \sqrt{3} cm$

Solution

(a) Given, $\angle Q P R=50^{\circ}$

We know that, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

$\therefore$ $\angle O P R=90^{\circ}$

$\Rightarrow$ $\angle O P Q+\angle Q P R=90^{\circ}$

[from figure]

$\Rightarrow$ $\angle O P Q=90^{\circ}-50^{\circ}=40^{\circ}$

$[\because \angle Q P R=50^{\circ}]$

Now,

$O P=O Q=$ Radius of circle

$ \therefore \quad \angle O Q P=\angle O P Q=40^{\circ} $

In $\triangle O P Q$

[since, angles opposite to equal sides are equal]

[since, sum of angles of a triangle $=180^{\circ}$ ]

$\Rightarrow$ $ \angle O+\angle P+\angle Q=180^{\circ} $

$.+40^{\circ}) \quad[\because \angle P=40^{\circ}=\angle Q]$

$ \begin{aligned} \angle O = 180^{\circ}-(40^{\circ}+40^{\circ}) \\ & =180^{\circ}-80^{\circ}=100^{\circ} \end{aligned} $

Solution

(a) Given, $P A$ and $P B$ are tangent lines.

[since, tangent at any point of a circle is perpendicular to the radius through the point of contact]

$ \begin{matrix} \therefore & \angle P A O = 90^{\circ} \\ \Rightarrow & \angle P A B+\angle B A O = 90^{\circ} \\ \Rightarrow & 65^{\circ}+\angle B A O = 90^{\circ} \\ \Rightarrow & \angle B A O = 90^{\circ}-65^{\circ}=25^{\circ} \end{matrix} $

Solution

(d) Let $P$ be an external point and a pair of tangents is drawn from point $P$ and angle between these two tangents is $60^{\circ}$.

Join $O A$ and $O P$.

Also, $O P$ is a bisector line of $\angle A P C$.

$ \begin{matrix} \therefore & \angle A P O=\angle C P O=30^{\circ} \\ \text{ Also, } & O A \perp A P \end{matrix} $

Tangent at any point of a circle is perpendicular to the radius through the point of contact.

In right angled $\triangle O A P$,

$ \begin{aligned} \tan 30^{\circ} = \dfrac{O A}{A P}=\dfrac{3}{A P} \\ \dfrac{1}{\sqrt{3}} = \dfrac{3}{A P} \\ A P = 3 \sqrt{3} cm \end{aligned} $

$ \Rightarrow \quad \dfrac{1}{\sqrt{3}}=\dfrac{3}{A P} $

$\Rightarrow$ Hence, the length of each tangent is $3 \sqrt{3} cm$.

Solution

(b) Given, $A B | P R$

Solution

False

Since a chord $A B$ subtends an angle of $60^{\circ}$ at the centre of a circle.

i.e.,

$\angle A O B=60^{\circ}$

As $\quad O A=O B=$ Radius of the circle

$\therefore \quad \angle O A B=\angle O B A=60^{\circ}$

The tangent at points $A$ and $B$ is drawn, which intersect at $C$.

We know, $O A \perp A C$ and $O B \perp B C$.

$\therefore \quad \angle O A C=90^{\circ}, \angle O B C=90^{\circ}$

$\Rightarrow \quad \angle O A B+\angle B A C=90^{\circ}$

and $\quad \angle O B A+\angle A B C=90^{\circ}$

$\Rightarrow \quad \angle B A C=90^{\circ}-60^{\circ}=30^{\circ}$

and $\quad \angle A B C=90^{\circ}-60^{\circ}=30^{\circ}$

In $\triangle A B C, \quad \angle B A C+\angle C B A+\angle A C B=180^{\circ}$

$\Rightarrow \quad \angle A C B=180^{\circ}-(30^{\circ}+30^{\circ})=120^{\circ}$

Solution

False

Because the length of tangent from an external point $P$ on a circle may or may not be greater than the radius of the circle.

Solution

True

$P T$ is a tangent drawn from external point $P$. Join $O T$.

$\because \quad O T \perp P T$

So, OPT is a right angled triangle formed.

In right angled triangle, hypotenuse is always greater than any of the two sides of the triangle. $\therefore$ or $O P>P T$ $P T<O P$

Solution

True

This may be possible only when both tangent lines coincide or are parallel to each other.

Solution

True

From point $P$, two tangents are drawn.

Given,

$O T=a$

Also, line $O P$ bisects the $\angle R P T$.

$\therefore$ Also,

In right angled $\triangle O T P$,

$\angle T P O=\angle R P O=45^{\circ}$

$O T \perp P T$

$\Rightarrow \quad \dfrac{1}{\sqrt{2}}=\dfrac{a}{O P} \Rightarrow O P=a \sqrt{2}$

Solution

False

From point $P$, two tangents are drawn.

Given,

$ O T=a $

Also, line $O P$ bisects the $\angle R P T$.

$ \begin{matrix} \therefore & \angle T P O=\angle R P O=30^{\circ} \\ \text{ Also, } & O T \perp P T \end{matrix} $

In right angled $\triangle O T P$,

$ \begin{matrix}& \sin 30^{\circ} = \dfrac{O T}{O P} \\ \Rightarrow & \dfrac{1}{2} = \dfrac{a}{O P} \\ \Rightarrow & O P = 2 a \end{matrix} $

Solution

True

Let $E A F$ be tangent to the circumcircle of $\triangle A B C$.

To prove

Here,

$\Rightarrow$ $E A F | B C$ $ \angle E A B=\angle A B C $

$ A B=A C $

$ \angle A B C=\angle A C B $

[angle between tangent and is chord equal to angle made by chord in the alternate segment] $\therefore$ Also, $\quad \angle E A B=\angle B C A$

From Eqs. (i) and (ii), we get

$\Rightarrow \quad E A F | B C$

Solution

False

Given that $P Q$ is any line segment and $S_1, S_2, S_3, S_4, \ldots$ circles are touch a line segment $P Q$ at a point $A$. Let the centres of the circles $S_1, S_2, S_3, S_4, \ldots$ be $C_1, C_2, C_3, C_4, \ldots$ respectively.

To prove Centres of these circles lie on the perpendicular bisector of $P Q$.

Now, joining each centre of the circles to the point $A$ on the line segment $P Q$ by a line segment i.e., $C_1 A, C_2 A, C_3 A, C_4 A$,… so on.

We know that, if we draw a line from the centre of a circle to its tangent line, then the line is always perpendicular to the tangent line. But it not bisect the line segment $P Q$.

So,

Since, each circle is passing through a point $A$. Therefore, all the line segments $C_1 A, C_2 A, C_3 A, C_4 A, \ldots$, so on are coincident.

So, centre of each circle lies on the perpendicular line of $P Q$ but they do not lie on the perpendicular bisector of $P Q$.

Hence, a number of circles touch a given line segment $P Q$ at a point $A$, then their centres lie

Solution

True

We draw two circle with centres $C_1$ and $C_2$ passing through the end points $P$ and $Q$ of a line segment $P Q$. We know, that perpendicular bisectors of a chord of a circle always passes through the centre of circle.

Thus, perpendicular bisector of $P Q$ passes through $C_1$ and $C_2$. Similarly, all the circle passing through $P Q$ will have their centre on perpendiculars bisectors of $P Q$.

Solution

True

To Prove, $B C=B D$

Join $B C$ and $O C$.

Given,

$\Rightarrow$ $\angle B A C=30^{\circ}$

$\angle B C D=30^{\circ}$

[angle between tangent and chord is equal to angle made by chord in the alternate

$ \begin{aligned} & \therefore \quad \angle A C D=\angle A C O+\angle O C D=30^{\circ}+90^{\circ}=120^{\circ} \\ & {[\because O C \perp C D \text{ and } O A=O C=\text{ radius } \Rightarrow \angle O A C=\angle O C A=30^{\circ}]} \\ & \text{ In } \triangle A C D \\ & \angle C A D+\angle A C D+\angle A D C=180^{\circ} \\ & \Rightarrow \quad 30^{\circ}+120^{\circ}+\angle A D C=180^{\circ} \\ & \Rightarrow \quad \angle A D C=180^{\circ}-(30^{\circ}+120^{\circ})=30^{\circ} \\ & \text{ Now, in } \triangle B C D \\ & \angle B C D=\angle B D C=30^{\circ} \\ & B C=B D \end{aligned} $

[since, sides opposite to equal angles are equal]

Solution

Let $C_1$ and $C_2$ be the two circles having same centre $O$. $A C$ is a chord which touches the $C_1$ at point $D$.

Solution

Given Two tangents $P Q$ and $P R$ are drawn from an external point to a circle with centre $O$.

To prove $Q O R P$ is a cyclic quadrilateral. proof Since, $P R$ and $P Q$ are tangents.

So,

$O R \perp P R$ and $O Q \perp P Q$

[since, if we drawn a line from centre of a circle to its tangent line. Then, the line always

$ \begin{matrix} \therefore & \angle O R P=\angle O Q P=90^{\circ} \\ \text{ Hence, } & \angle O R P+\angle O Q P=180^{\circ} \end{matrix} $

perpendicular to the tangent line]

So, QOPR is cyclic quadrilateral.

[If sum of opposite angles is quadrilateral in $180^{\circ}$, then the quadrilateral is cyclic]

Hence proved.

Solution

Given Two tangents $P Q$ and $P R$ are drawn from an external point $P$ to a circle with centre $O$.

To prove Centre of a circle touching two intersecting lines lies on the angle bisector of the lines.

In $\angle R P Q$.

Construction Join $O R$, and $O Q$.

In $\triangle P O R$ and $\triangle P O Q$

$ \angle P R O=\angle P Q O=90^{\circ} $

[tangent at any point of a circle is perpendicular to the radius through the point of contact]

$ O R=O Q $

[radii of some circle]

Since, $O P$ is common. $\therefore$ $\triangle P R O \cong \triangle P Q O$ Hence, $\angle R P O=\angle Q P O$

[RHS]

[by $CPCT]$

Thus, $O$ lies on angle bisecter of $P R$ and $P Q$.

Hence proved.

Solution

Two tangents $B D$ and $B C$ are drawn from an external point $B$.

To prove

Given,

Join $O C, O D$ and $B O$.

Since, $B C$ and $B D$ are tangents.

$\therefore \quad O C \perp B C$ and $O D \perp B D$

We know, $O B$ is a angle bisector of $\angle D B C$.

$\therefore \quad \angle O B C=\angle D B O=60^{\circ}$

In right angled $\triangle O B C$,

$\cos 60^{\circ}=\dfrac{B C}{O B}$

$\Rightarrow \quad \dfrac{1}{2}=\dfrac{B C}{O B}$

$\Rightarrow \quad O B=2 B C$

Also, $\quad B C=B D$

[tangent drawn from internal point to circle are equal]

$\therefore \quad O B=B C+B C$

$\Rightarrow \quad O B=B C+B D$

Solution

Given $A B$ and $C D$ are common tangent to two circles of unequal radius To prove $A B=C D$

Construction Produce $A B$ and $C D$, to intersect at $P$.

Proof $\quad P A=P C$

[the length of tangents drawn from an internal point to a circle are equal]

Also,

$ P B=P D $

[the lengths of tangents drawn from an internal point to a circle are equal]

$ \therefore \quad P A-P B=P C-P D $

$ A B=C D $

Hence proved.

Solution

Given $A B$ and $C D$ are tangents to two circles of equal radii. To prove

$A B=C D$

Construction Join $O A, O C, O^{\prime} B$ and $O^{\prime} D$

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

[tangent at any point of a circle is perpendicular to radius through the point of contact] Thus, $A C$ is a straight line.

Also,

$\therefore$

Similarly, $B D$ is a straight line and

Also,

In quadrilateral $A B C D$, and $A B C D$ is a rectangle Hence,

$ \begin{gathered} \angle O A B+\angle O C D=180^{\circ} \\ A B | C D \end{gathered} $

$ \begin{aligned} \angle O^{\prime} B A = \angle O^{\prime} D C=90^{\circ} \\ A C = B D \quad \text{ [radii of two circles are equal] } \\ \angle A = \angle B=\angle C=\angle D=90^{\circ} \\ A C = B D \end{aligned} $

$A B=C D \quad$ [opposite sides of rectangle are equal]

Solution

Given Common tangents $A B$ and $C D$ to two circles intersecting at $E$.

To prove $A$

$B=C D$

Proof

$$ \begin{equation*} E A=E C \tag{i} \end{equation*} $$

[the lengths of tangents drawn from an internal point to a circle are equal]

$$ \begin{equation*} E B=E D \tag{ii} \end{equation*} $$

On adding Eqs. (i) and (ii), we get

$ \begin{aligned} E A+E B = E C+E D \\ A B = C D \end{aligned} $

Hence proved.

Solution

Given Chord $P Q$ is parallel to tangent at $R$.

To prove $R$ bisects the arc $P R Q$

$\angle 1=\angle 3$

[angle between tangent and chord is equal to angle made by chord in alternate segment]

$ \begin{matrix} \therefore & \angle 2=\angle 3 \\ \Rightarrow & P R=Q R \\ \Rightarrow & P R=Q R \end{matrix} $

Solution

To prove $\angle 1=\angle 2$, let $P Q$ be a chord of the circle. Tangents are drawn at the points $R$ and Q.

Let $P$ be another point on the circle, then, join $P Q$ and $P R$.

Since, at point $Q$, there is a tangent.

$\therefore \quad \angle 2=\angle P \quad$ [angles in alternate segments are equal]

Since, at point $R$, there is a tangent.

$\therefore \quad \angle 1=\angle P \quad$ [angles in alternate segments are equal]

$\therefore \quad \angle 1=\angle 2=\angle P$

Hence proved.

Solution

Given, $A B$ is a diameter of the circle.

A tangent is drawn from point $A$. Draw a chord $C D$ parallel to the tangent MAN.

So, $C D$ is a chord of the circle and $O A$ is a radius of the circle.

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

[tangent at any point of a circle is perpendicular to the radius through the point of contact] $\angle C E O=\angle M A O$ $\angle C E O=90^{\circ}$

[corresponding angles]

$\therefore$

Thus, $O E$ bisects $C D, \quad$ [perpendicular from centre of circle to chord bisects the chord] Similarly, the diameter $A B$ bisects all. Chords which are parallel to the tangent at the point $A$.

Solution

Given $A$ hexagon $A B C D E F$ circumscribe a circle.

To prove $A B+C D+E F=B C+D E+F A$

Proof $A B+C D+E F=(A Q+Q B)+(C S+S D)+(E U+U F)$

$=A P+B R+C R+D T+E T+F P$

$=(A P+F P)+(B R+C R)+(D T+E T)$

$ \begin{aligned} A B+C D+E F = A F+B C+D E \\ A Q = A P \\ Q B = B R \\ C S = C R \\ D S = D T \\ E U = E T \end{aligned} $

[tangents drawn from an external point to a circle are equal]

Hence proved.

Solution

A circle is inscribed in the $\triangle A B C$, which touches the $B C, C A$ and $A B$.

Given,

$ B C=a, C A=b \text{ and } A B=c $

By using the property, tangents are drawn from an external point to the circle are equal in length.

$ \begin{aligned} & \therefore \quad B D=B F=x \quad \text{ [say] } \\ & \text{ ond } \\ & \text{ Now, } \quad B C+C A+A B=a+b+c \\ & \Rightarrow \quad(B D+D C)+(C E+E A)+(A F+F B)=a+b+c \\ & \Rightarrow \quad(x+y)+(y+z)+(z+x)=a+b+c \\ & \Rightarrow \quad 2(x+y+z)=2 s \\ & \Rightarrow \quad s=x+y+z \\ & \Rightarrow \quad x=s-(y+z) \\ & \Rightarrow \quad B D=s-b \quad[\because b=A E+E C=z+y] \end{aligned} $

Solution

Two tangents $P A$ and $P B$ are drawn to a circle with centre $O$ from an external point $P$.

Solution

Since, $A C$ is a diameter line, so angle in semi-circle makes an angle $90^{\circ}$. $\therefore$ In $\triangle A B C$, $\angle A B C=90^{\circ}$ $\angle C A B+\angle A B C+\angle A C B=180^{\circ}$ [by property]

$[\because.$ sum of all interior angles of any triangle is $.180^{\circ}]$

$\Rightarrow \quad \angle C A B+\angle A C B=180^{\circ}-90^{\circ}=90^{\circ}$

Since, diameter of a circle is perpendicular to the tangent.

i.e.

$\therefore$ $C A \perp A T$

$\angle C A T=90^{\circ}$

$\Rightarrow$ $\angle C A B+\angle B A T=90^{\circ}$

From Eqs. (i) and (ii),

$ \begin{aligned} \Rightarrow & \angle C A B+\angle A C B = \angle C A B+\angle B A T \\ \angle A C B = \angle B A T \end{aligned} $

Hence proved.

Solution

Here, two circles are of radii $O P=3 cm$ and $P O^{\prime}=4 cm$.

These two circles intersect at $P$ and $Q$.

Here, $O P$ and $P O^{\prime}$ are two tangents drawn at point $P$.

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

[tangent at any point of circle is perpendicular to radius through the point of contact]

Join $O O^{\prime}$ and $P N$.

In right angled $\triangle O P O^{\prime}$,

$ \begin{aligned} & (O O^{\prime})^{2}=(O P)^{2}+(P O^{\prime})^{2} \quad \text{ [by Pythagoras theorem] } \\ & \text{ i.e., } \quad(\text{ Hypotenuse })^{2}=(\text{ Base })^{2}+(\text{ Perpendicular })^{2} \\ & =(3)^{2}+(4)^{2}=25 \\ & \Rightarrow \quad O O^{\prime}=5 cm \\ & \text{ Also, } \quad P N \perp O O^{\prime} \end{aligned} $

$ (O P)^{2}=(O N)^{2}+(N P)^{2} \quad[\text{ by Pythagoras theorem }] $

$$ \begin{equation*} \Rightarrow \quad(N P)^{2}=3^{2}-x^{2}=9-x^{2} \tag{i} \end{equation*} $$

and in right angled $\triangle P N O^{\prime}$,

$ \begin{aligned} & (P O^{\prime})^{2}=(P N)^{2}+(N O^{\prime})^{2} \quad[\text{ by Pythagoras theorem }] \\ & \Rightarrow \quad(4)^{2}=(P N)^{2}+(5-x)^{2} \\ & \Rightarrow \quad(P N)^{2}=16-(5-x)^{2} \end{aligned} $

From Eqs. (i) and (ii),

$ 9-x^{2}=16-(5-x)^{2} $

$\Rightarrow \quad 7+x^{2}-(25+x^{2}-10 x)=0$

$ \Rightarrow \quad 10 x=18 $

$ \therefore \quad x=1.8 $

Again, in right angled $\triangle O P N$,

$ \begin{matrix} \Rightarrow & 3^{2}=(1.8)^{2}+(N P)^{2} \\ \Rightarrow & (N P)^{2} = 9-3.24=5.76 \\ \therefore & (N P) = 2.4 \\ \therefore \text{ Length of common chord, } & P Q = 2 P N=2 \times 2.4=4.8 cm \end{matrix} $

[by Pythagoras theorem]

Solution

Let $O$ be the centre of the given circle. Suppose, the tangent at $P$ meets $B C$ at $Q$. Join $B P$.

Proof

[tangent at any point of circle is perpendicular to radius through the point of contact]

$\therefore \ln \triangle A B C, \quad \begin{aligned} \angle 1+\angle 5 = 90^{\circ} \\ \angle 3 = \angle 1\end{aligned} \quad$ [angle sum property, $\angle A B C=90^{\circ}$ ]

[angle between tangent and the chord equals angle made by the chord in alternate segment]

$\therefore$ $\angle 3+\angle 5=90^{\circ}$ $\ldots$ (i)

Also, $\angle A P B=90^{\circ}$ [angle in semi-circle]

$\Rightarrow$ $\angle 3+\angle 4=90^{\circ}$ $[\angle A P B+\angle B P C=180^{\circ}.$, linear pair]

From Eqs. (i) and (ii), we get

$\Rightarrow$ $\angle 3+\angle 5=\angle 3+\angle 4$

$\Rightarrow$ $\angle 5=\angle 4$

Also, $P Q=Q C$ [sides opposite to equal angles are equal]

$\Rightarrow$ $Q P=Q B$

$\Rightarrow$ $[$ tangents drawn from an internal point to a circle are equal]

$Q B=Q C$

Solution

$P Q$ and $P R$ are two tangents drawn from an external point $P$.

$\therefore \quad P Q=P R$

[the lengths of tangents drawn from an external point to a circle are equal]

$\Rightarrow$ $\angle P Q R=\angle Q R P$

[angles opposite to equal sides are equal]

Now, in $\triangle P Q R \quad \angle P Q R+\angle Q R P+\angle R P Q=180^{\circ}$

[sum of all interior angles of any triangle is $180^{\circ}$ ]

$\Rightarrow \quad \angle P Q R+\angle P Q R+30^{\circ}=180^{\circ}$

$\Rightarrow \quad 2 \angle P Q R=180^{\circ}-30^{\circ}$

$\Rightarrow \quad \angle P Q R=\dfrac{180^{\circ}-30^{\circ}}{2}=75^{\circ}$

Since, $\quad S R | Q P$

$\therefore \quad \angle S R Q=\angle R Q P=75^{\circ} \quad$ [alternate interior angles]

Also, $\quad \angle P Q R=\angle Q S R=75^{\circ} \quad$ [by alternate segment theorem]

In $\triangle Q R S, \quad \angle Q+\angle R+\angle S=180^{\circ}$

[sum of all interior angles of any triangle is $180^{\circ}$ ]

$ \begin{matrix} \Rightarrow & \angle Q = 180^{\circ}-(75^{\circ}+75^{\circ}) \\ & =30^{\circ} \\ \therefore & \angle R Q S = 30^{\circ} \end{matrix} $

Solution

A circle is drawn with centre $O$ and $A B$ is a diameter.

$A C$ is a chord such that $\angle B A C=30^{\circ}$.

Given $A B$ is $\alpha$ diameter and $A C$ is a chord of circle with certre $O, \angle B A C=30^{\circ}$.

To prove

$\ln \triangle A B C$,

$ \begin{aligned} & \angle A+\angle B+\angle C=180^{\circ} \\ & 30^{\circ}+\angle B+90^{\circ}=180^{\circ} \\ & \Rightarrow \\ & \text{ Also, } \\ & \Rightarrow \\ & \angle C B A+\angle C B D=180^{\circ} \\ & \angle C B D=180^{\circ}-60^{\circ}-120^{\circ} \\ & {[\because \angle C B A=60^{\circ}]} \\ & \Rightarrow \quad 120^{\circ}+\angle B D C+30^{\circ}=180^{\circ} \\ & \Rightarrow \quad \angle B D C=30^{\circ} \\ & \text{ From Eqs. (i) and (ii), } \\ & \angle B C D=\angle B D C \\ & B C=B D \end{aligned} $

Solution

Let mid-point of an $arc A M B$ be $M$ and $T M T^{\prime}$ be the tangent to the circle.

Join $A B, A M$ and $M B$.

Since,

$ \begin{aligned} arc A M = arc M B \\ \text{ Chord } A M = \text{ Chord } M B \\ A M = M B \\ \angle M A B = \angle M B A \end{aligned} $

$\Rightarrow$ In $\triangle A M B$,

$\Rightarrow$ [equal sides corresponding to the equal angle] …(i)

Since, $T M T^{\prime}$ is a tangent line. $\therefore$ $\angle A M T=\angle M B A$ [angles in alternate segments are equal] $=\angle M A B$ [from Eq. (i)]

But $\angle A M T$ and $\angle M A B$ are alternate angles, which is possible only when

$A B | T M T^{\prime}$

Hence, the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

Hence proved.

Solution

Joint $A O, O C$ and $O^{\prime} D, O^{\prime} B$. Now, in $\triangle E O^{\prime} D$ and $\triangle E O^{\prime} B$,

$$ \begin{align*} O^{\prime} D = O^{\prime} B \tag{radius}\\ O^{\prime} E = O^{\prime} E \tag{commonside}\\ E D = E B \end{align*} $$

[since, tangents drawn from an external point to the circle are equal in length]

$ \begin{aligned} & \therefore \quad \triangle E O^{\prime} D \cong \triangle E O^{\prime} B \quad \text{ [by SSS congruence rule] } \\ & \Rightarrow \quad \angle O^{\prime} E D=\angle O^{\prime} E B \end{aligned} $

$O^{\prime} E$ is the angle bisector of $\angle D E B$.

Similarly, $O E$ is the angle bisector of $\angle A E C$.

Now, in quadrilateral $D E B O^{\prime}$,

$ \angle O^{\prime} D E=\angle O^{\prime} B E=90^{\circ} $

$ \Rightarrow \quad \angle O^{\prime} D E+\angle O^{\prime} B E=180^{\circ} $

$\therefore \quad \angle D E B+\angle D O^{\prime} B=180^{\circ}$ [since, $D E B O^{\prime}$ is cyclic quadrilateral] … (ii)

Since, $A B$ is a straight line.

$ \begin{aligned} & \therefore \quad \angle A E D+\angle D E B=180^{\circ} \\ & \Rightarrow \quad \angle A E D+180^{\circ}-\angle D O^{\prime} B=180^{\circ} \\ & \Rightarrow \\ & \angle A E D=\angle D O^{\prime} B \\ & \angle A E D=\angle A O C \\ & \angle D E B=180^{\circ}-\angle D O^{\prime} B \end{aligned} $

Divided by 2 on both sides, we get

$$ \begin{align*} \Rightarrow \dfrac{1}{2} \angle D E B = 90^{\circ}-\dfrac{1}{2} \angle D O^{\prime} B \\ \angle D E O^{\prime} = 90^{\circ}-\dfrac{1}{2} \angle D O^{\prime} B \tag{v} \end{align*} $$

[since, $O^{\prime} E$ is the angle bisector of $\angle D E B$ i.e., $\dfrac{1}{2} \angle D E B=\angle D E O^{\prime}$ ]

Similarly,

$ \angle A E C=180^{\circ}-\angle A O C $

Divided by 2 on both sides, we get

$$ \begin{matrix} \Rightarrow & \dfrac{1}{2} \angle A E C=90^{\circ}-\dfrac{1}{2} \angle A O C \\ \Rightarrow \quad \angle A E O=90^{\circ}-\dfrac{1}{2} \angle A O C \tag{vi} \end{matrix} $$

[since, $O E$ is the angle bisector of $\angle A E C$ i.e., $\dfrac{1}{2} \angle A E C=\angle A E O$ ]

Now, $\angle A E D+\angle D E O^{\prime}+\angle A E O=\angle A E D+90^{\circ}-\dfrac{1}{2} \angle D O^{\prime} B+90^{\circ}-\dfrac{1}{2} \angle A O C$

$ \begin{aligned} & =\angle A E D+180^{\circ}-\dfrac{1}{2}(\angle D O^{\prime} B+\angle A O C) \\ & =\angle A E D+180^{\circ}-\dfrac{1}{2}(\angle A E D+\angle A E D) \quad[\text{ from Eqs. (iii) and (iv) }] \\ & =\angle A E D+180^{\circ}-\dfrac{1}{2}(2 \times \angle A E D) \\ & =\angle A E D+180^{\circ}-\angle A E D=180^{\circ} \end{aligned} $

$ \therefore \quad \angle A E O+\angle A E D+\angle D E O^{\prime}=180^{\circ} $

So, $O E O$ ’ is straight line.

Hence, $O, E$ and $O^{\prime}$ are collinear.

Hence proved.

Solution

Given, $O T=13 cm$ and $O P=5 cm$

Since, if we drawn a line from the centre to the tangent of the circle. It is always perpendicular to the tangent i.e., $O P \perp P T$.

In right angled $\triangle O P T, \quad O T^{2}=O P^{2}+P T^{2}$

$ \begin{matrix} \Rightarrow & P T^{2}=(13)^{2}-(5)^{2}=169-25=144 \\ \Rightarrow & P T=12 cm \end{matrix} $

$\therefore$
$\therefore$ $P A=A E$ and $Q B=E B$
$O T=13 cm$ …(iii)

$\therefore$ $E T=O T-O E$ $[\therefore O E=5 cm=$ radius $]$

$\Rightarrow$ $E T=13-5$

$ [\text{ by Pythagoras theorem, }(\text{ hypotenuse })^{2}=(\text{ base })^{2}+(\text{ perpendicular })^{2}] $

Since, the length of pair of tangents from an external point $T$ is equal.

$$ \begin{matrix} \therefore & Q T=12 cm \\ \text{ Now, } & T A=P T-P A \\ \Rightarrow & T A=12-P A \\ \text{ and } & T B=Q T-Q B \\ \Rightarrow & T B=12-Q B \tag{ii} \end{matrix} $$

Again, using the property, length of pair of tangents from an external point is equal.

Since, $A B$ is a tangent and $O E$ is the radius.

$ \therefore $

$\Rightarrow \quad \angle O E A=90^{\circ}$

$\therefore \quad \angle A E T=180^{\circ}-\angle O E A$

$\Rightarrow$ $ \angle A E T=90^{\circ} $

Now, in right angled $\triangle A E T$,

$ (A T)^{2}=(A E)^{2}+(E T)^{2} \quad[\text{ by Pythagoras theorem }] $

$ \begin{matrix} \Rightarrow & (P T-P A)^{2}=(A E)^{2}+(8)^{2} & \\ \Rightarrow & (12-P A)^{2}=(P A)^{2}+(8)^{2} & \text{ [from Eq. (iii)] } \\ \Rightarrow & 144+(P A)^{2}-24 \cdot P A=(P A)^{2}+64 & \\ \Rightarrow & 24 \cdot P A=80 & \\ \Rightarrow & P A=\dfrac{10}{3} cm & \\ \therefore & A E=\dfrac{10}{3} cm & \text{ [from Eq. (iii)] } \end{matrix} $

Join OQ.

Similarly

$ \begin{aligned} B E = \dfrac{10}{3} cm \\ A B = A E+E B \\ & =\dfrac{10}{3}+\dfrac{10}{3} \\ & =\dfrac{20}{3} cm \end{aligned} $

Hence,

[linear pair]

[from Eq. (iii)]

Hence,the required length $A B$ is $\dfrac{20}{3} cm$.

Solution

Here, $A B$ is a diameter of the circle from point $C$ and a tangent is drawn which meets at a point $P$.

Join OC. Here, OC is radius.

Since, tangent at any point of a circle is perpendicular to the radius through point of contact circle.

Now

$ \begin{gathered} O C \perp P C \\ \angle P C A=110^{\circ} \end{gathered} $

$ \begin{aligned} \Rightarrow & \angle P C O+\angle O C A = 110^{\circ} \\ \Rightarrow & 90^{\circ}+\angle O C A = 110^{\circ} \\ \Rightarrow & \angle O C A = 20^{\circ} \\ \therefore & O C = O A=\text{ Radius of circle } \\ \Rightarrow & \angle O C A = \angle O A C=20^{\circ} \end{aligned} $

[since, two sides are equal, then their opposite angles are equal]

Since, $P C$ is a tangent, so

$ \angle B C P=\angle C A B=20^{\circ} $

[angles in a alternate segment are equal]

In $\triangle P B C, \quad \angle P+\angle C+\angle A=180^{\circ}$

$ \begin{aligned} \angle P = 180^{\circ}-(\angle C+\angle A) \\ & =180^{\circ}-(110^{\circ}+20^{\circ}) \\ & =180^{\circ}-130^{\circ}=50^{\circ} \end{aligned} $

$\ln \triangle P B C$

$\angle B P C+\angle P C B+\angle P B C=180^{\circ}$

$\Rightarrow \quad 50^{\circ}+20^{\circ}+\angle P B C=180^{\circ}$

$\Rightarrow \quad \angle P B C=180^{\circ}-70^{\circ}$

$\Rightarrow \quad \angle P B C=110^{\circ}$

Since, $A P B$ is a straight line.

$ \begin{aligned} & \therefore \quad \angle P B C+\angle C B A=180^{\circ} \\ & \Rightarrow \quad \angle C B A=180^{\circ}-110^{\circ}=70^{\circ} \end{aligned} $

Solution

In a circle, $\triangle A B C$ is inscribed.

Join $O B, O C$ and $O A$.

Conside $\triangle A B O$ and $\triangle A C O$

$ \begin{aligned} & A B=A C \quad \text{[given]}\\ & B O=C O \quad \text{[radii of same circle]} \end{aligned} $

$A O$ is common. $\therefore$ $\triangle A B O \cong \triangle A C O$

$\therefore$ $\triangle A M B \cong \triangle A M C$ [by SAS congruence rule]

$\angle A M B=\angle A M C$ [CPCT]

Also,

$ \begin{aligned} & \angle A M B+\angle A M C = 180^{\circ} \\ \Rightarrow & \angle A M B = 90^{\circ} \end{aligned} $ [linear pair]

$\angle 1=\angle 2$ [by SSS congruence rule]

$ \begin{aligned} & A B=A C \\ & \angle 1=\angle 2 \end{aligned} $

[CPOT] Now, in $\triangle A B M$ and $\triangle A C M$, $A M$ is common. [given] [proved above]

We know that a perpendicular from centre of circle bisects the chord.

So, $O A$ is perpendicular bisector of $B C$. Let $A M=x$, then $O M=9-x$ $[\because O A=$ radius $=9 cm]$

In right angled $\triangle A M C, \quad A C^{2}=A M^{2}+M C^{2}$

i.e.,

$(\text{ Hypotenuse })^{2}=(\text{ Base })^{2}+(\text{ Perpendicular })^{2}$

$$ \begin{equation*} M C^{2}=6^{2}-x^{2} \tag{i} \end{equation*} $$

and in right $\triangle O M C, \quad O C^{2}=O M^{2}+M C^{2} \quad$ [by Pythagoras theorem]

$\Rightarrow$ $M C^{2}=9^{2}-(9-x)^{2}$

From Eqs. (i) and (ii),

In right angled $\triangle A B M$,

From eqs. (i) and (ii), $\quad$ $ 6^{2}-x^{2}=9^{2}-(9-x)^{2} $

$ \begin{array}{ll} \Rightarrow & 36-x^{2} = 81-(81+x^{2}-18 x) &\\ \Rightarrow & 36 = 18 x \Rightarrow x=2 \\ \therefore & A M = x=2 \\ \text{In right angled } \triangle ABM, & A B^{2} = B M^{2}+A M^{2} & \text{ [by Pythagoras theorem] } \\ 6^{2} = B M^{2}+2^{2} \\ \Rightarrow & B M^{2} = 36-4=32 \\ \Rightarrow & B M = 4 \sqrt{2} \\ \therefore B C = 2 B M=8 \sqrt{2} cm \\ \therefore \text{ Area of } \triangle A B C = \dfrac{1}{2} \times \text{ Base } \times \text{ Height } \\ & =\dfrac{1}{2} \times B C \times A M \\ & =\dfrac{1}{2} \times 8 \sqrt{2} \times 2=8 \sqrt{2} cm^{2} \end{array} $

[by Pythagoras theorem]

Hence, the required area of $\triangle A B C$ is $8 \sqrt{2} cm^{2}$.

Solution

Given Two tangents are drawn from an external point $A$ to the circle with centre $O$,

$ O A=13 cm $

Tangent $B C$ is drawn at a point $R$. radius of circle equals $5 cm$.

To find perimeter of $\triangle A B C$. Proof

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

[tangent at any point of a circle is perpendicular to the radius through the point of contact]

$ \begin{aligned} & \therefore \quad O A^{2}=O P^{2}+P A^{2} \quad \text{ [by Pythagoras theorm] } \\ & \Rightarrow \quad P A^{2}=144=12^{2} \\ & \Rightarrow \quad P A=12 cm \\ & \text{ Now, } \quad \text{ perimeter of } \triangle A B C=A B+B C+C A \\ & =(A B+B R)+(R C+C A) \\ & =A B+B P+C Q+C A \end{aligned} $

$[A P=A Q$ tangent from internal point to a circle are equal $]$

Hence, the perimeter of $\triangle A B C=24 cm$.