Circle — Page 7 | JEE Mathematics

Q61

Choose the incorrect statement about the two circles whose equations are given below : x2 + y2

-

10x

-

10y + 41 = 0 and x2 + y2

-

16x

-

10y + 80 = 0

A Distance between two centres is the average of radii of both the circles.

B Both circles pass through the centre of each other.

C Circles have two intersection points.

D Both circle's centers lie inside region of one another.

Correct Answer

Option D

Solution

S1 $\equiv$ x2 + y2 $-$ 10x $-$ 10y + 41 = 0 Centre C1 $\equiv$ (5, 5), radius r1 = 3 S2 $\equiv$ x2 + y2 $-$ 16x $-$ 10y + 80 = 0 Centre C2 $\equiv$ (8, 5), radius r2 = 3 Distance between centres = 3 Hence both circles pass through the centre of each other, have two intersection point and distance between two centres in average of radii of both the circles.

Hence, option (d) is the incorrect statement.

Q62

Let the tangent to the circle x2 + y2 = 25 at the point R(3, 4) meet x-axis and y-axis at points P and Q, respectively. If r is the radius of the circle passing through the origin O and having centre at the incentre of the triangle OPQ, then r2 is equal to :

A

{{585} \over {66}}

B

{{625} \over {72}}

C

{{529} \over {64}}

D

{{125} \over {72}}

Correct Answer

Option B

Solution

Given equation of circle x2 + y2 = 25 $\therefore$ Tangent equation at (3, 4) T : 3x + 4y = 25 Incentre of

\Delta

OPQ.

I = \left( {{{{{25} \over 4} \times {{25} \over 3}} \over {{{25} \over 3} + {{25} \over 4} + {{125} \over {12}}}},{{{{25} \over 3} \times {{25} \over 4}} \over {{{25} \over 3} + {{25} \over 4} + {{125} \over {12}}}}} \right)

$\therefore$

I = \left( {{{625} \over {75 + 100 + 125}},{{625} \over {75 + 100 + 125}}} \right) = \left( {{{25} \over {12}},{{25} \over {12}}} \right)

$\because$ Distance from origin to incenter is r. $\therefore$

{r^2} = {\left( {{{25} \over {12}}} \right)^2} + {\left( {{{25} \over {12}}} \right)^2} = {{625} \over {72}}

Therefore, the correct answer is (B).

Q63

Two tangents are drawn from a point P to the circle x2 + y2

-

2x

-

4y + 4 = 0, such that the angle between these tangents is

{\tan ^{ - 1}}\left( {{{12} \over 5}} \right)

, where

{\tan ^{ - 1}}\left( {{{12} \over 5}} \right)

\in

(0,

\pi

). If the centre of the circle is denoted by C and these tangents touch the circle at points A and B, then the ratio of the areas of

\Delta

PAB and

\Delta

CAB is :

A 3 : 1

B 9 : 4

C 2 : 1

D 11 : 4

Correct Answer

Option B

Solution

Let $\theta$ = tan $-$ 1

\left( {{{12} \over 5}} \right)

$\Rightarrow$ tan $\theta$ =

{{{12} \over 5}}

$\Rightarrow$

{{2\tan {\theta \over 2}} \over {1 - {{\tan }^2}{\theta \over 2}}} = {{12} \over 5}

\Rightarrow \tan {\theta \over 2} = {2 \over 3}

\Rightarrow \sin {\theta \over 2} = {2 \over {\sqrt 3 }}

and

\cos {\theta \over 2} = {3 \over {\sqrt {13} }}

In

\Delta

CAP,

\tan {\theta \over 2} = {1 \over {AP}}

\Rightarrow AP = {3 \over 2}

In

\Delta

APM,

\sin {\theta \over 2} = {{AM} \over {AP}},\cos {\theta \over 2} = {{PM} \over {AP}}

\Rightarrow AM = {3 \over {\sqrt {13} }}

\Rightarrow PM = {9 \over {2\sqrt {13} }}

$\therefore$

AB = {6 \over {\sqrt {13} }}

$\therefore$ Area of

\Delta PAB = {1 \over 2} \times AB \times PM

= {1 \over 2} \times {6 \over {\sqrt {13} }} \times {9 \over {2\sqrt {13} }} = {{27} \over {16}}

Now,

\phi = 90^\circ - {\theta \over 2}

In

\Delta

CAM,

\cos \phi = {{CM} \over {CA}}

\Rightarrow CM = 1.\cos \left( {{\pi \over 2} - {\theta \over 2}} \right)

= 1.\sin {\theta \over 2} = {2 \over {\sqrt {13} }}

$\therefore$ Area of

\Delta CAB = {1 \over 2} \times AB \times CM

= {1 \over 2} \times {6 \over {\sqrt {13} }} \times {2 \over {\sqrt {13} }} = {6 \over {13}}

$\therefore$

{{Area\,of\,\Delta PAB} \over {Area\,of\,\Delta CAB}} = {{27/26} \over {6/13}} = {9 \over 4}

Therefore, the correct answer is (2).

Q64

Choose the correct statement about two circles whose equations are given below : x2 + y2

-

10x

-

10y + 41 = 0 x2 + y2

-

22x

-

10y + 137 = 0

A circles have same centre

B circles have no meeting point

C circles have only one meeting point

D circles have two meeting points

Correct Answer

Option C

Solution

Let

{S_1}:{x^2} + {y^2} - 10x - 10y + 41 = 0

\Rightarrow {(x - 5)^2} + {(y - 5)^2} = 9

Centre

({C_1}) = (5,5)

Radius r1 = 3

{S_2}:{x^2} + {y^2} - 22x - 10y + 137 = 0

\Rightarrow {(x - 11)^2} + {(y - 5)^2} = 9

Centre

({C_2}) = (11,5)

Radius r2 = 3 distance

({C_1}{C_2}) = \sqrt {{{(5 - 11)}^2} + {{(5 - 5)}^2}}

distance

({C_1}{C_2}) = 6

$\because$

{r_1} + {r_2} = 3 + 3 = 6

$\therefore$ circles touch externally Hence, circle have only one meeting point.

Q65

Let S1 : x2 + y2 = 9 and S2 : (x

-

2)2 + y2 = 1. Then the locus of center of a variable circle S which touches S1 internally and S2 externally always passes through the points :

A

\left( {{1 \over 2}, \pm {{\sqrt 5 } \over 2}} \right)

B (1,

\pm

2)

C

\left( {2, \pm {3 \over 2}} \right)

D (0,

\pm

\sqrt 3

)

Correct Answer

Option C

Solution

S1 : x2 + y2 = 9 ; C1 (0, 0), r1 = 3 S2 : (x $-$ 2)2 + y2 = 1 ; C2 (2, 0), r2 = 1 Image Let the variable circle S and its radius is r units.

Here S and S1 touches internally $\therefore$ Distance between center, S + S1 = PC1 = 3 $-$ r Here S and S2 touches externally $\therefore$ Distance between center, S + S2 = PC2 = 1 + r $\therefore$ PC1 + PC2 = 4 > C1 C2 So locus is ellipse whose focii are C1 & C2 and major axis is 2a = 4 and 2ae = C1C2 = 2 $\Rightarrow$

e = {1 \over 2}

$\Rightarrow$

{b^2} = 4\left( {1 - {1 \over 4}} \right) = 3

Centre of ellipse is midpoint of C1 & C2 is (1, 0) Equation of ellipse is

{{{{(x - 1)}^2}} \over {{2^2}}} + {{{y^2}} \over {{{\left( {\sqrt 3 } \right)}^2}}} = 1

Now by cross checking the option

\left( {2, \pm {3 \over 2}} \right)

satisfied it.

Q66

Let r1 and r2 be the radii of the largest and smallest circles, respectively, which pass through the point (

-

4, 1) and having their centres on the circumference of the circle x2 + y2 + 2x + 4y

-

4 = 0. If

{{{r_1}} \over {{r_2}}} = a + b\sqrt 2

, then a + b is equal to :

A 3

B 11

C 5

D 7

Correct Answer

Option C

Solution

Centre of smallest circle is A Centre of largest circle is B

{r_2} = |CP - CA| = 3\sqrt 2 - 3

{r_1} = CP - CB = 3\sqrt 2 + 3

{{{r_1}} \over {{r_2}}} = {{3\sqrt 2 + 3} \over {3\sqrt 2 - 3}} = {{{{(3\sqrt 2 + 3)}^2}} \over 9} = {(\sqrt 2 + 1)^2} = 3 + 2\sqrt 2

a = 3, b = 2

Q67

Let the circle S : 36x2 + 36y2

-

108x + 120y + C = 0 be such that it neither intersects nor touches the co-ordinate axes. If the point of intersection of the lines, x

-

2y = 4 and 2x

-

y = 5 lies inside the circle S, then :

A

{{25} \over 9} < C < {{13} \over 3}

B 100 < C < 165

C 81 < C < 156

D 100 < C < 156

Correct Answer

Option D

Solution

S : 36x2 + 36y2 $-$ 108x + 120y + C = 0 $\Rightarrow$ x2 + y2 $-$ 3x +

{{10} \over 3}

y +

{C \over {36}}

= 0 Centre

\equiv ( - g, - f) \equiv \left( {{3 \over 2},{{ - 10} \over 6}} \right)

radius =

r = \sqrt {{9 \over 4} + {{100} \over {36}} - {C \over {36}}}

Now,

\Rightarrow r < {3 \over 2}

\Rightarrow {9 \over 4} + {{100} \over {36}} - {C \over {36}} < {9 \over 4}

$\Rightarrow$ C > 100 ......

(1) Now, point of intersection of x $-$ 2y = 4 and 2x $-$ y = 5 is (2, $-$ 1), which lies inside the circle S.

$\therefore$ S(2, $-$ 1) < 0 $\Rightarrow$ (2)2 + ( $-$ 1)2 $-$ 3(2) +

{{10} \over 3}

( $-$ 1) +

{C \over {36}}

< 0 $\Rightarrow$ 4 + 1 $-$ 6 $-$

{{10} \over 3}

+

{C \over {36}}

< 0 C < 156 ..... (2) From (1) & (2) 100 < C < 156 Ans.

Q68

Consider a circle C which touches the y-axis at (0, 6) and cuts off an intercept

6\sqrt 5

on the x-axis. Then the radius of the circle C is equal to :

A

\sqrt {53}

B 9

C 8

D

\sqrt {82}

Correct Answer

Option B

Solution

r = \sqrt {{6^2} + {{\left( {3 + \sqrt 5 } \right)}^2}}

= \sqrt {36 + 45} = 9

Q69

Two tangents are drawn from the point P(

-

1, 1) to the circle x2 + y2

-

2x

-

6y + 6 = 0. If these tangents touch the circle at points A and B, and if D is a point on the circle such that length of the segments AB and AD are equal, then the area of the triangle ABD is equal to :

A 2

B

(3\sqrt 2 + 2)

C 4

D

3(\sqrt 2 - 1)

Correct Answer

Option C

Solution

\Delta ABD = {1 \over 2} \times 2 \times 4 = 4

Q70

Let P and Q be two distinct points on a circle which has center at C(2, 3) and which passes through origin O. If OC is perpendicular to both the line segments CP and CQ, then the set {P, Q} is equal to :

A {(4, 0), (0, 6)}

B

\{ (2 + 2\sqrt 2 ,3 - \sqrt 5 ),(2 - 2\sqrt 2 ,3 + \sqrt 5 )\}

C

\{ (2 + 2\sqrt 2 ,3 + \sqrt 5 ),(2 - 2\sqrt 2 ,3 - \sqrt 5 )\}

D {(

-

1, 5), (5, 1)}

Correct Answer

Option D

Solution

\tan \theta = - {2 \over 3}

Using symmetric from of line

P,Q:\left( {2 \pm \sqrt {13} \cos \theta ,3 \pm \sqrt {13} \sin \theta } \right)

\left( {2 \pm \sqrt {13} .\left( { - {3 \over {\sqrt {13} }}} \right),3 \pm \sqrt {13} \left( {{2 \over {\sqrt {13} }}} \right)} \right)

( $-$ 1, 5) & (5, 1)