Exercise 10.2 (Revised) - Chapter 10 - Circles - Ncert Solutions class 10 - Maths

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Chapter 10: Circles - NCERT Solutions for Class 10 Maths

Ex 10.2 Question 1.

From a point $Q$, the length of the tangent to a circle is $24 \mathrm{~cm}$ and the distance of $Q$ 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) $24.5 \mathrm{~cm}$

Answer.

(A)
$
\because \angle \mathrm{OPQ}=90^{\circ}
$
[The tangent at any point of a circle is $\qquad$ to the radius
through the point of contact]

$\therefore$ In right triangle $\mathrm{OPQ}$,
$
O Q^2=O P^2+P Q^2
$
[By Pythagoras theorem]
$
\begin{aligned}
& \Rightarrow(25)^2=O P^2+(24)^2 \\
& \Rightarrow 625=O P^2+576
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow O P^2=625-576=49 \\
& \Rightarrow O P=7 \mathrm{~cm}
\end{aligned}
$
Ex 10.2 Question 2.

In figure, if TP and TQ are the two tangents to a circle with centre 0 so that $\angle P O Q=$ $110^{\circ}$, then $\angle$ PTQ is equal to:

(A) $60^{\circ}$
(B) $70^{\circ}$
(C) $80^{\circ}$
(D) $90^{\circ}$

Answer.

(B)
$
\angle \mathrm{POQ}=110^{\circ}, \angle \mathrm{OPT}=90^{\circ} \text { and } \angle \mathrm{OQT}=90^{\circ}
$
[The tangent at any point of a circle is $\perp$ to the radius through the point of contact]
In quadrilateral OPTQ
$
\angle \mathrm{POQ}+\angle \mathrm{OPT}+\angle \mathrm{OQT}+\angle \mathrm{PTQ}=360^{\circ}
$
[Angle sum property of quadrilateral]
$
\begin{aligned}
& \Rightarrow 110^{\circ}+90^{\circ}+90^{\circ}+\angle \mathrm{PTQ}=360^{\circ} \\
& \Rightarrow 290^{\circ}+\angle \mathrm{PTQ}=360^{\circ} \\
& \Rightarrow \angle \mathrm{PTQ}=360^{\circ}-290^{\circ} \\
& \Rightarrow \angle \mathrm{PTQ}=70^{\circ}
\end{aligned}
$

Ex 10.2 Question 3.

If tangents PA and PB from a point $P$ to a circle with centre $O$ 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) $80^{\circ}$

Answer.

(A)
$
\because \angle \mathrm{OAP}=90^{\circ}
$
[The tangent at any point of a circle is $\perp$ to the radius
through the point of contact]

$
\angle \mathrm{OPA}=\frac{1}{2} \angle \mathrm{BPA}=\frac{1}{2} \times 80^{\circ}=40^{\circ}
$
[Centre lies on the bisector of the
angle between the two tangents]
In $\triangle \mathrm{OPA}$,
$
\angle \mathrm{OAP}+\angle \mathrm{OPA}+\angle \mathrm{POA}=180^{\circ}
$
[Angle sum property of a triangle]
$
\begin{aligned}
& \Rightarrow 90^{\circ}+40^{\circ}+\angle \mathrm{POA}=180^{\circ} \\
& \Rightarrow 130^{\circ}+\angle \mathrm{POA}=180^{\circ} \\
& \Rightarrow \angle \mathrm{POA}=50^{\circ}
\end{aligned}
$

Ex 10.2 Question 4.

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

Answer.

Given: $\mathrm{PQ}$ is a diameter of a circle with centre 0 .
The lines $A B$ and $C D$ are the tangents at $P$ and $Q$ respectively.

To Prove: $A B \| C D$
Proof: Since $\mathrm{AB}$ is a tangent to the circle at $\mathrm{P}$ and $\mathrm{OP}$ is the radius through the point of contact.
$\therefore \angle \mathrm{OPA}=90^{\circ}$
[The tangent at any point of a circle is $\perp$ to the radius through the point of contact]
$\because \mathrm{CD}$ is a tangent to the circle at $\mathrm{Q}$ and $\mathrm{OQ}$ is the radius through the point of contact.
$\therefore \angle \mathrm{OQD}=90^{\circ}$
[The tangent at any point of a circle is $\perp$ to the radius through the point of contact]
From eq. (i) and (ii), $\angle \mathrm{OPA}=\angle \mathrm{OQD}$
But these form a pair of equal alternate angles also,
$\therefore A B|| C D$
Ex 10.2 Question 5.

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

Answer.

Let $\mathrm{AB}$ be the tangent drawn at the point $\mathrm{P}$ on the circle with $\mathrm{O}$.

If possible, let $\mathrm{PQ}$ be perpendicular to $\mathrm{AB}$, not passing through $\mathrm{O}$. Join OP.

Since tangnet at a point to a circle is perpendicular to the radius through the point.
Therefore, $\mathrm{AB} \perp \mathrm{OP} \quad \Rightarrow \quad \angle \mathrm{OPB}=90^{\circ}$

Also, $\angle \mathrm{QPB}=90^{\circ}$
[By construction]

Therefore, $\angle \mathrm{QPB}=\angle \mathrm{OPB}$, which is not possible as a part cannot be equal to whole.
Thus, it contradicts our supposition.
Hence, the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Ex 10.2 Question 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.

Answer.

We know that the tangent at any point of a circle is $\perp$ to the radius through the point of contact.

$
\begin{aligned}
& \therefore \angle \mathrm{OPA}=90^{\circ} \\
& \therefore O A^2=O P^2+A P^2
\end{aligned}
$
[By Pythagoras theorem]
$
\begin{aligned}
& \Rightarrow(5)^2=(\mathrm{OP})^2+(4)^2 \\
& \Rightarrow 25=(\mathrm{OP})^2+16 \\
& \Rightarrow O P^2=9 \\
& \Rightarrow O P=3 \mathrm{~cm}
\end{aligned}
$
Ex 10.2 Question 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.

Answer.

Let O be the common centre of the two concentric circles.

Let $\mathrm{AB}$ be a chord of the larger circle which touches the smaller circle at $\mathrm{P}$.
Join OP and OA.
Then, $\angle \mathrm{OPA}=90^{\circ}$
[The tangent at any point of a circle is $\perp$ to the radius through the point of contact]
$
\therefore \mathrm{OA}^2=\mathrm{OP}^2+\mathrm{AP}^2
$
[By Pythagoras theorem]
$
\begin{aligned}
& \Rightarrow(5)^2=(3)^2+A P^2 \\
& \Rightarrow 25=9+A P^2 \\
& \Rightarrow A P^2=16 \\
& \Rightarrow A P=4 \mathrm{~cm}
\end{aligned}
$

Since the perpendicular from the centre of a circle to a chord bisects the chord, therefore
$
\begin{aligned}
& \mathrm{AP}=\mathrm{BP}=4 \mathrm{~cm} \\
& \Rightarrow \mathrm{AB}=\mathrm{AP}+\mathrm{BP} \\
& =\mathrm{AP}+\mathrm{AP}=2 \mathrm{AP} \\
& =2 \times 4=8 \mathrm{~cm}
\end{aligned}
$
Ex 10.2 Question 8.

A quadrilateral $\mathrm{ABCD}$ is drawn to circumscribe a circle (see figure). Prove that:
$
A B+C D=A D+B C
$

Answer.

We know that the tangents from an external point to a circle are equal.
$
\therefore A P=A S
$
$
\mathrm{BP}=\mathrm{BQ}
$
$
C R=C Q
$
$
\text { DR }=\text { DS }
$

On adding eq. (i), (ii), (iii) and (iv), we get
$
\begin{aligned}
& (\mathrm{AP}+\mathrm{BP})+(\mathrm{CR}+\mathrm{DR}) \\
& =(\mathrm{AS}+\mathrm{BQ})+(\mathrm{CQ}+\mathrm{DS}) \\
& \Rightarrow \mathrm{AB}+\mathrm{CD}=(\mathrm{AS}+\mathrm{DS})+(\mathrm{BQ}+\mathrm{CQ}) \\
& \Rightarrow \mathrm{AB}+\mathrm{CD}=\mathrm{AD}+\mathrm{BC}
\end{aligned}
$
Ex 10.2 Question 9.

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

Answer.

Given: In figure, $\mathrm{XY}$ and $\mathrm{X}^{\prime} \mathrm{Y}^{\prime}$ are two parallel tangents to a circle with centre $\mathrm{O}$ and another tangent $\mathrm{AB}$ with point of contact $\mathrm{C}$ intersecting $\mathrm{XY}$ at $\mathrm{A}$ and $\mathrm{X}^{\prime} \mathrm{Y}^{\prime}$ at $\mathrm{B}$.

To Prove: $\angle \mathrm{AOB}=90^{\circ}$
Construction: Join OC
Proof: $\angle \mathrm{OPA}=90^{\circ} \ldots \ldots \ldots .$. (i)

$
\angle \mathrm{OCA}=90^{\circ}
$
[Tangent at any point of a circle is $\perp$ to the radius through the point of contact]

In right angled triangles OPA and OCA,
$
\begin{aligned}
& \angle \mathrm{OPA}=\angle \mathrm{OCA}=90^{\circ} \\
& \mathrm{OA}=\mathrm{OA} \text { [Common] }
\end{aligned}
$
$\mathrm{AP}=\mathrm{AC}$ [Tangents from an external

point to a circle are equal]
$
\therefore \triangle \mathrm{OPA} \cong \triangle \mathrm{OCA}
$
[RHS congruence criterion]
$\therefore \angle \mathrm{OAP}=\angle \mathrm{OAC}[$ By C.P.C.T.]
$\Rightarrow \angle \mathrm{OAC}=\frac{1}{2} \angle \mathrm{PAB}$
Similarly, $\angle \mathrm{OBQ}=\angle \mathrm{OBC}$
$\Rightarrow \angle \mathrm{OBC}=\frac{1}{2} \angle \mathrm{QBA}$
$\because$ XY \| X'Y' and a transversal AB intersects them.
$\therefore \angle \mathrm{PAB}+\angle \mathrm{QBA}=180^{\circ}$
[Sum of the consecutive interior angles on the same side of the transversal is $180^{\circ}$ ]
$\Rightarrow \frac{1}{2} \angle \mathrm{PAB}+\frac{1}{2} \angle \mathrm{QBA}$
$=\frac{1}{2} \times 180^{\circ}$
$\Rightarrow \angle \mathrm{OAC}+\angle \mathrm{OBC}=90^{\circ}$
[From eq. (iii) \& (iv)]
In $\triangle \mathrm{AOB}$,
$\angle \mathrm{OAC}+\angle \mathrm{OBC}+\angle \mathrm{AOB}=180^{\circ}$
[Angel sum property of a triangle]

$\begin{aligned}
& \Rightarrow 90^{\circ}+\angle A O B=180^{\circ}[\text { From eq. (v) }] \\
& \Rightarrow \angle A O B=90^{\circ}
\end{aligned}$

Hence proved.
Ex 10.2 Question 10.

Prove that the angel 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.

$\angle \mathrm{OAP}=90^{\circ}$ $\qquad$
$\angle \mathrm{OBP}=90^{\circ}$. $\qquad$ (ii)

[Tangent at any point of a circle is $\perp$ to
the radius through the point of contact]
$\because$ OAPB is quadrilateral.
$\therefore \angle \mathrm{APB}+\angle \mathrm{AOB}+\angle \mathrm{OAP}+\angle \mathrm{OBP}=360^{\circ}$
[Angle sum property of a quadrilateral]
$\Rightarrow \angle \mathrm{APB}+\angle \mathrm{AOB}+90^{\circ}+90^{\circ}=360^{\circ}$
[From eq. (i) \& (ii)]
$\Rightarrow \angle \mathrm{APB}+\angle \mathrm{AOB}=180^{\circ}$
$\therefore \angle \mathrm{APB}$ and $\angle \mathrm{AOB}$ are supplementary.
Ex 10.2 Question 11.

Prove that the parallelogram circumscribing a circle is a rhombus.

Answer.

Given: ABCD is a parallelogram circumscribing a circle.

To Prove: $\mathrm{ABCD}$ is a rhombus.

Proof: Since, the tangents from an external point to a circle are equal.
$
\therefore \mathrm{AP}=\mathrm{AS}
$

$
\begin{aligned}
& B P=B Q \\
& C R=C Q \\
& D R=D S
\end{aligned}
$
$\qquad$
$\qquad$
$\qquad$
On adding eq. (i), (ii), (iii) and (iv), we get
$
\begin{aligned}
& (\mathrm{AP}+\mathrm{BP})+(\mathrm{CR}+\mathrm{DR}) \\
& =(\mathrm{AS}+\mathrm{BQ})+(\mathrm{CQ}+\mathrm{DS}) \\
& \Rightarrow \mathrm{AB}+\mathrm{CD}=(\mathrm{AS}+\mathrm{DS})+(\mathrm{BQ}+\mathrm{CQ}) \\
& \Rightarrow \mathrm{AB}+\mathrm{CD}=\mathrm{AD}+\mathrm{BC} \\
& \Rightarrow \mathrm{AB}+\mathrm{AB}=\mathrm{AD}+\mathrm{AD}
\end{aligned}
$
[Opposite sides of \| gm are equal]
$
\begin{aligned}
& \Rightarrow 2 \mathrm{AB}=2 \mathrm{AD} \\
& \Rightarrow \mathrm{AB}=\mathrm{AD}
\end{aligned}
$

But $\mathrm{AB}=\mathrm{CD}$ and $\mathrm{AD}=\mathrm{BC}$

[Opposite sides of \| gm]
$\therefore \mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{AD}$
$\therefore$ Parallelogram $\mathrm{ABCD}$ is a rhombus.
Ex 10.2 Question 12.

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

Answer.

Join OE and OF. Also join OA, OB and OC. 

Since $\mathrm{BD}=8 \mathrm{~cm}$
$
\therefore \mathrm{BE}=8 \mathrm{~cm}
$

[Tangents from an external point to a circle are equal]
Since $C D=6 \mathrm{~cm}$
$\therefore \mathrm{CF}=6 \mathrm{~cm}$
[Tangents from an external point to a circle are equal]
Let $\mathrm{AE}=\mathrm{AF}=x$
Since $O D=O E=O F=4 \mathrm{~cm}$
[Radii of a circle are equal]
$
\begin{aligned}
& \therefore \text { Semi-perimeter of } \Delta \mathrm{ABC}=\frac{(x+6)+(x+8)+(6+8)}{2}=(x+14) \mathrm{cm} \\
& \therefore \text { Area of } \Delta \mathrm{ABC}=\sqrt{s(s-a)(s-b)(s-c)} \\
& =\sqrt{(x+14)(x+14-14)(x+14-\overline{x+8})(x+14-\overline{x+6})} \\
& =\sqrt{(x+14)(x)(6)(8)} \mathrm{cm}^2
\end{aligned}
$

Now, Area of $\triangle \mathrm{ABC}=$ Area of $\triangle \mathrm{OBC}+$ Area of $\triangle O C A+$ Area of $\triangle O A B$
$
\begin{aligned}
& \Rightarrow \sqrt{(x+14)(x)(6)(8)} \\
& =\frac{(6+8) 4}{2}+\frac{(x+6) 4}{2}+\frac{(x+8) 4}{2} \\
& \Rightarrow \sqrt{(x+14)(x)(6)(8)} \\
& =28+2 x+12+2 x+16
\end{aligned}
$

$\begin{aligned}
& \Rightarrow \sqrt{(x+14)(x)(6)(8)}=4 x+56 \\
& \Rightarrow \sqrt{(x+14)(x)(6)(8)}=4(x+14)
\end{aligned}$

Squaring both sides,
$
\begin{aligned}
& (x+14)(x)(6)(8)=16(x+14)^2 \\
& \Rightarrow 3 x=x+14 \\
& \Rightarrow 2 x=14 \\
& \Rightarrow x=7 \\
& \therefore \mathrm{AB}=x+8=7+8=15 \mathrm{~cm}
\end{aligned}
$

And $\mathrm{AC}=x+6=7+6=13 \mathrm{~cm}$
Ex 10.2 Question 13.

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

Answer.

Given: $\mathrm{ABCD}$ is a quadrilateral circumscribing a circle whose centre is $\mathrm{O}$.
To prove: (i) $\angle \mathrm{AOB}+\angle \mathrm{COD}=180^{\circ}$ (ii) $\angle \mathrm{BOC}+\angle \mathrm{AOD}=180^{\circ}$
Construction: Join OP, OQ, OR and OS.

Proof: Since tangents from an external point to a circle are equal.
$
\begin{aligned}
& \therefore \mathrm{AP}=\mathrm{AS}, \\
& \mathrm{BP}=\mathrm{BQ} \ldots
\end{aligned}
$
$
C Q=C R
$

$
D R=D S
$

In $\triangle \mathrm{OBP}$ and $\triangle \mathrm{OBQ}$,
$\mathrm{OP}=\mathrm{OQ}$ [Radii of the same circle]
$\mathrm{OB}=\mathrm{OB}$ [Common]
$\mathrm{BP}=\mathrm{BQ}[$ [From eq. (i)]
$\therefore \Delta \mathrm{OPB} \cong \Delta \mathrm{OBQ}$ [By SSS congruence criterion]
$\therefore \angle 1=\angle 2[$ By c.P.C.T.]
Similarly, $\angle 3=\angle 4, \angle 5=\angle 6, \angle 7=\angle 8$
Since, the sum of all the angles round a point is equal to $360^{\circ}$.
$
\begin{aligned}
& \therefore \angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8=360^{\circ} \\
& \Rightarrow \angle 1+\angle 1+\angle 4+\angle 4+\angle 5+\angle 5+\angle 8+\angle 8=360^{\circ} \\
& \Rightarrow 2(\angle 1+\angle 4+\angle 5+\angle 8)=360^{\circ} \\
& \Rightarrow \angle 1+\angle 4+\angle 5+\angle 8=180^{\circ} \\
& \Rightarrow(\angle 1+\angle 5)+(\angle 4+\angle 8)=180^{\circ} \\
& \Rightarrow \angle \mathrm{AOB}+\angle \mathrm{COD}=180^{\circ}
\end{aligned}
$

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