NCERT Solutions for Exercise 10.2 Class 10 Maths Chapter 10 - Circles

Circles Class 10 Chapter 10 Circles Exercise: 10.2

Q1 From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the center is 25 cm. The radius of the circle is

(A) 7 cm

(B) 12 cm

(C) 15 cm

(D) 24.5 cm

Answer:

The correct option is (A) = 7 cm
Given that,
The length of the tangent (QT) is 24 cm and the length of OQ is 25 cm.
Suppose the length of the radius OT be $l$ cm.
We know that $\mathrm{\Delta }OTQ$ is a right angle triangle. So, by using Pythagoras theorem-

$$\begin{gathered} O{Q}^2=T{Q}^2+O{T}^2 \\ l=\sqrt{{25}^2-{24}^2} \\ OT=l=\sqrt{49} \end{gathered}$$

OT = 7 cm

Q2 In Fig. 10.11, if TP and TQ are the two tangents to a circle with center O so that $\mathrm{\angle }$ POQ = $110^{\circ }$ , then $\mathrm{\angle }$ PTQ is equal to

(A) ${60}^{\circ }$

(B) ${70}^{\circ }$

(C) ${80}^{\circ }$

(D) $90^{\circ }$

Answer:

The correct option is (b)

In figure, $\mathrm{\angle }POQ={110}^0$
Since POQT is quadrilateral. Therefore the sum of the opposite angles are 180

$$\begin{gathered} \Rightarrow \mathrm{\angle }PTQ+\mathrm{\angle }POQ={180}^0 \\ \Rightarrow \mathrm{\angle }PTQ={180}^0-\mathrm{\angle }POQ \end{gathered}$$
$={180}^0-{100}^0$
$={70}^0$

Q3 If tangents PA and PB from a point P to a circle with center O are inclined to each other at an angle of ${80}^{\circ }$ , then $\mathrm{\angle }$ POA is equal to

(A) 50°

(B) 60°

(C) 70°

(D) 80°

Answer:

The correct option is (A)

It is given that, tangent PA and PB from point P inclined at $\mathrm{\angle }APB={80}^0$
In triangle $\mathrm{\Delta }$ OAP and $\mathrm{\Delta }$ OBP
$\mathrm{\angle }OAP=\mathrm{\angle }OBP=90$
OA =OB (radii of the circle)
PA = PB (tangents of the circle)

Therefore, by SAS congruence
$\therefore \mathrm{\Delta }OAP\cong \mathrm{\Delta }OBP$

By CPCT, $\mathrm{\angle }OPA=\mathrm{\angle }OPB$
Now, $\mathrm{\angle }$ OPA = 80/4 = 40

In $\mathrm{\Delta }$ PAO,
$\mathrm{\angle }P+\mathrm{\angle }A+\mathrm{\angle }O=180$
$\mathrm{\angle }O=180-130$
= 50

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

Answer:

Let line $p$ and line $q$ are two tangents of a circle and AB is the diameter of the circle.
OA and OB are perpendicular to the tangents $p$ and $q$ respectively.
therefore,
$\mathrm{\angle }1=\mathrm{\angle }2={90}^0$

$\Rightarrow$ $p$ || $q$ { $\mathrm{\angle }$ $\because$ 1 & $\mathrm{\angle }$ 2 are alternate angles}

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

Answer:

In the above figure, the line AXB is the tangent to a circle with center O. Here, OX is the perpendicular to the tangent AXB ( $OX\perp AXB$ ) at point of contact X.
Therefore, we have,
$\mathrm{\angle }$ BXO + $\mathrm{\angle }$ YXB = ${90}^0+{90}^0={180}^0$

$\therefore$ OXY is a collinear
$\Rightarrow$ OX is passing through the center of the circle.

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

Answer:

Given that,
the length of the tangent from the point A (AP) is 4 cm and the length of OA is 5 cm.
Since $\mathrm{\angle }$ APO = 90 ⁰
Therefore, $\mathrm{\Delta }$ APO is a right-angle triangle. By using Pythagoras theorem;

$O{A}^2=A{P}^2+O{P}^2$
$5^2=4^2+O{P}^2$
$OP=\sqrt{25-16}=\sqrt{9}$
$OP=3cm$

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:

In the above figure, Pq is the chord to the larger circle, which is also tangent to a smaller circle at the point of contact R.
We have,
radius of the larger circle OP = OQ = 5 cm
radius of the small circle (OR) = 3 cm

OR $\perp$ PQ [since PQ is tangent to a smaller circle]

According to question,

In $\mathrm{\Delta }$ OPR and $\mathrm{\Delta }$ OQR
$\mathrm{\angle }$ PRO = $\mathrm{\angle }$ QRO {both ${90}^0$ }

OR = OR {common}
OP = OQ {both radii}

By RHS congruence $\mathrm{\Delta }$ OPR $\cong$ $\mathrm{\Delta }$ OQR
So, by CPCT
PR = RQ
Now, In $\mathrm{\Delta }$ OPR,
by using pythagoras theorem,
$PR=\sqrt{25-9}=\sqrt{16}$
PR = 4 cm
Hence, PQ = 2.PR = 8 cm

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

Answer:

To prove- AB + CD = AD + BC
Proof-
We have,
Since the length of the tangents drawn from an external point to a circle are equal
AP =AS .......(i)
BP = BQ.........(ii)
AS = AP...........(iii)
CR = CQ ...........(iv)

By adding all the equations, we get;

AP + BP +RD+ CR = AS +DS +BQ +CQ
$\Rightarrow$ (AP + BP) + (RD + CR) = (AS+DS)+(BQ + CQ)
$\Rightarrow$ AB + CD = AD + BC

Hence proved.

Q9 In Fig. 10.13, XY and X'Y'are two parallel tangents to a circle with center O and another tangent AB with a point of contact C intersecting XY at A and X'Y' at B. Prove that $\mathrm{\angle }$ AOB = 90°.

Answer:

To prove- $\mathrm{\angle }$ AOB = ${90}^0$
Proof-
In $\mathrm{\Delta }$ AOP and $\mathrm{\Delta }$ AOC,
OA =OA [Common]
OP = OC [Both radii]
AP =AC [tangents from external point A]
Therefore by SSS congruence, $\mathrm{\Delta }$ AOP $\cong$ $\mathrm{\Delta }$ AOC
and by CPCT, $\mathrm{\angle }$ PAO = $\mathrm{\angle }$ OAC
$\Rightarrow \mathrm{\angle }PAC=2\mathrm{\angle }OAC$ ..................(i)

Similarly, from $\mathrm{\Delta }$ OBC and $\mathrm{\Delta }$ OBQ, we get;
$\mathrm{\angle }$ QBC = 2. $\mathrm{\angle }$ OBC.............(ii)

Adding eq (1) and eq (2)

$\mathrm{\angle }$ PAC + $\mathrm{\angle }$ QBC = 180
2( $\mathrm{\angle }$ OBC + $\mathrm{\angle }$ OAC) = 180
( $\mathrm{\angle }$ OBC + $\mathrm{\angle }$ OAC) = 90

Now, in $\mathrm{\Delta }$ OAB,
Sum of interior angle is 180.
So, $\mathrm{\angle }$ OBC + $\mathrm{\angle }$ OAC + $\mathrm{\angle }$ AOB = 180
$\therefore$ $\mathrm{\angle }$ AOB = 90
hence proved.

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 center.

Answer:

To prove - $\mathrm{\angle }APB+\mathrm{\angle }AOB={180}^0$
Proof-
We have, PA and PB are two tangents, B and A are the point of contacts of the tangent to a circle. And $OA\perp PA$ , $OB\perp PB$ (since tangents and radius are perpendiculars)

According to question,
In quadrilateral PAOB,
$\mathrm{\angle }$ OAP + $\mathrm{\angle }$ APB + $\mathrm{\angle }$ PBO + $\mathrm{\angle }$ BOA = ${360}^0$
90 + $\mathrm{\angle }$ APB + 90 + $\mathrm{\angle }$ BOA = 360
$\mathrm{\angle }APB+\mathrm{\angle }AOB={180}^0$
Hence proved .

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

Answer:

To prove - the parallelogram circumscribing a circle is a rhombus
Proof-
ABCD is a parallelogram that circumscribes a circle with center O.
P, Q, R, S are the points of contacts on sides AB, BC, CD, and DA respectively
AB = CD .and AD = BC...........(i)
It is known that tangents drawn from an external point are equal in length.
RD = DS ...........(ii)
RC = QC...........(iii)
BP = BQ...........(iv)
AP = AS .............(v)
By adding eq (ii) to eq (v) we get;
(RD + RC) + (BP + AP) = (DS + AS) + (BQ + QC)
CD + AB = AD + BC
$\Rightarrow$ 2AB = 2AD [from equation (i)]
$\Rightarrow$ AB = AD
Now, AB = AD and AB = CD
$\therefore$ AB = AD = CD = BC

Hence ABCD is a rhombus.

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 (see Fig. 10.14). Find the sides AB and AC.

Answer:

Consider the above figure. Assume center O touches the sides AB and AC of the triangle at point E and F respectively.

Let the length of AE is x.

Now in $△ABC$ ,

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

$BE=BD=6$ (tangents on the circle from point B)

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

Now AB = AE + EB

$$\begin{gathered} AB=AE+EB \\ =>AB=x+8 \\ BC=BD+DC \\ =>BC=8+6=14 \\ CA=CF+FA \\ =>CA=6+x \end{gathered}$$

Now
$$\begin{gathered} s=(AB+BC+CA)/2 \\ =>s=(x+8+14+6+x)/2 \\ =>s=(2x+28)/2 \\ =>s=x+14 \end{gathered}$$

Area of triangle $△ABC$

$$\begin{gathered} =\sqrt{s\ast (s-a)\ast (s-b)\ast (s-c)} \\ =\sqrt{(14+x)\ast [(14+x)-14]\ast [(14+x)-(6+x)]\ast [(14+x)-(8+x)]} \\ =4\sqrt{3(14x+x^2)} \end{gathered}$$
Now the area of $△OBC$

$$\begin{gathered} =(1/2)\ast OD\ast BC \\ =(1/2)\ast 4\ast 14 \\ =56/2=28 \end{gathered}$$
Area of $△OCA$

$$\begin{gathered} =(1/2)\ast OF\ast AC \\ =(1/2)\ast 4\ast (6+x) \\ =2(6+x) \\ =12+2x \end{gathered}$$

Area of $△OAB$

$$\begin{gathered} =(1/2)\ast OE\ast AB \\ =(1/2)\ast 4\ast (8+x) \\ =2(8+x) \\ =16+2x \end{gathered}$$

Now Area of the $△ABC$ = Area of $△OBC$ + Area of $△OCA$ + Area of $△OAB$

$$\begin{gathered} =>4\sqrt{3x(14+x)}=28+12+2x+16+2x \\ =>4\sqrt{3x(14+x)}=56+4x \\ =>4\sqrt{3x(14+x)}=4(14+x) \\ =>\sqrt{3x(14+x)}]=14+x \end{gathered}$$

On squaring both the side, we get

$$\begin{gathered} 3x(14+x)=(14+x{)}^2 \\ =>3x=14+x(14+x=0=>x=-14isnotpossible) \\ =>3x-x=14 \\ =>2x=14 \\ =>x=14/2 \\ =>x=7 \end{gathered}$$

Hence

AB = x + 8

=> AB = 7+8

=> AB = 15

AC = 6 + x

=> AC = 6 + 7

=> AC = 13

Answer- AB = 15 and AC = 13

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

Answer:

Given- ABCD is a quadrilateral circumscribing a circle. P, Q, R, S are the point of contact on sides AB, BC, CD, and DA respectively.

To prove-
$$\begin{gathered} \mathrm{\angle }AOB+\mathrm{\angle }COD={180}^0 \\ \mathrm{\angle }AOD+\mathrm{\angle }BOC={180}^0 \end{gathered}$$
Proof -
Join OP, OQ, OR and OS
In triangle $\mathrm{\Delta }$ DOS and $\mathrm{\Delta }$ DOR,
OD =OD [common]
OS = OR [radii of same circle]
DR = DS [length of tangents drawn from an external point are equal ]
By SSS congruency, $\mathrm{\Delta }$ DOS $\cong$ $\mathrm{\Delta }$ DOR,
and by CPCT, $\mathrm{\angle }$ DOS = $\mathrm{\angle }$ DOR
$\mathrm{\angle }c=\mathrm{\angle }d$ .............(i)

Similarily,
$$\begin{gathered} \mathrm{\angle }a=\mathrm{\angle }b \\ \mathrm{\angle }e=\mathrm{\angle }f \\ \mathrm{\angle }g=\mathrm{\angle }h \end{gathered}$$ ...............(2, 3, 4)

$\therefore 2(\mathrm{\angle }a+\mathrm{\angle }e+\mathrm{\angle }h+\mathrm{\angle }d)={360}^0$
$$\begin{gathered} (\mathrm{\angle }a+\mathrm{\angle }e)+(\mathrm{\angle }h+\mathrm{\angle }d)={180}^0 \\ \mathrm{\angle }AOB+\mathrm{\angle }DOC={180}^0 \end{gathered}$$
SImilarily, $\mathrm{\angle }AOD+\mathrm{\angle }BOC={180}^0$

Hence proved.