Circle — Page 8 | JEE Mathematics

Q71

Let

A = \{ (x,y) \in R \times R|2{x^2} + 2{y^2} - 2x - 2y = 1\}

,

B = \{ (x,y) \in R \times R|4{x^2} + 4{y^2} - 16y + 7 = 0\}

and

C = \{ (x,y) \in R \times R|{x^2} + {y^2} - 4x - 2y + 5 \le {r^2}\}

. Then the minimum value of |r| such that

A \cup B \subseteq C

is equal to

A

{{3 + \sqrt {10} } \over 2}

B

{{2 + \sqrt {10} } \over 2}

C

{{3 + 2\sqrt 5 } \over 2}

D

1 + \sqrt 5

Correct Answer

Option C

Solution

{S_1}:{x^2} + {y^2} - x - y - {1 \over 2} = 0;{C_1}\left( {{1 \over 2},{1 \over 2}} \right)

{r_1} = \sqrt {{1 \over 4} + {1 \over 4} + {1 \over 2}} = 1

{S_2}:{x^2} + {y^2} - 4y + {7 \over 4} = 0;{C_2}:(0,2)

{r_2} = \sqrt {4 - {7 \over 4}} = {3 \over 2}

{S_3} = {x^2} + {y^2} - 4x - 2y + 5 - {r^2} = 0

C3 (2, 1)

{r_3} = \sqrt {4 + 1 - 5 + {r^2}} = |r|

{C_1}{C_3} = \sqrt {{5 \over 2}}

\sqrt {{5 \over 2}} \le |r - 1| \Rightarrow \left. \begin{array}{ll}r \le 1 + \sqrt {{5 \over 2}} \\ r \ge {3 \over 2} + \sqrt 5 \end{array} \right\}

{C_2}{C_3} = \sqrt 5 \le \left| {r - {3 \over 2}} \right|

\left. \begin{array}{ll}r - {3 \over 2} \ge \sqrt 5 \\ r - {3 \over 2} \le - \sqrt 5 \end{array} \right\}

Q72

If a line along a chord of the circle 4x2 + 4y2 + 120x + 675 = 0, passes through the point (

-

30, 0) and is tangent to the parabola y2 = 30x, then the length of this chord is :

A 5

B 7

C 5

{\sqrt 3 }

D 3

{\sqrt 5 }

Correct Answer

Option D

Solution

Equation of tangent to y2 = 30 x y = mx +

{{30} \over {4m}}

Pass thru ( $-$ 30, 0) : a = $-$ 30m +

{{30} \over {4m}}

$\Rightarrow$ m2 = 1/4 $\Rightarrow$ m =

{1 \over 2}

or m = $-$

{1 \over 2}

At m =

{1 \over 2}

: y =

{x \over 2}

+ 15 $\Rightarrow$ x $-$ 2y + 30 = 0 lAB = 2

\sqrt {{R^2} - {P^2}}

= 2

\sqrt {{{225} \over 4} - {{225} \over 5}}

$\Rightarrow$ lAB = 30 .

\sqrt {{1 \over {20}}}

=

{{{15} \over {\sqrt 5 }}}

= 3

{\sqrt 5 }

Q73

A circle C touches the line x = 2y at the point (2, 1) and intersects the circle C1 : x2 + y2 + 2y

-

5 = 0 at two points P and Q such that PQ is a diameter of C1. Then the diameter of C is :

A

7\sqrt 5

B 15

C

\sqrt {285}

D

4\sqrt {15}

Correct Answer

Option A

Solution

(x $-$ 2)2 + (y $-$ 1)2 + $\lambda$ (x $-$ 2y) = 0 C : x2 + y2 + x( $\lambda$ $-$ 4) + y( $-$ 2 $-$ 2 $\lambda$ ) + 5 = 0 C1 : x2 + y2 + 2y $-$ 5 = 0 S1 $-$ S2 = 0 (Equation of PQ) ( $\lambda$ $-$ 4)x $-$ (2 $\lambda$ + 4)y + 10 = 0 Passes through (0, $-$ 1) $\Rightarrow$ $\lambda$ = $-$ 7 C : x2 + y2 $-$ 11x + 12y + 5 = 0 =

{{\sqrt {245} } \over 4}

Diameter =

7\sqrt 5

Q74

Let the tangent to the circle C1 : x2 + y2 = 2 at the point M(

-

1, 1) intersect the circle C2 : (x

-

3)2 + (y

-

2)2 = 5, at two distinct points A and B. If the tangents to C2 at the points A and B intersect at N, then the area of the triangle ANB is equal to :

A

{1 \over 2}

B

{2 \over 3}

C

{1 \over 6}

D

{5 \over 3}

Correct Answer

Option C

Solution

Equation of tangent at point M is

T = 0

\Rightarrow x{x_1} + y{y_1} = 2

\Rightarrow - x + y = 2

\Rightarrow y = x + 2

Putting this value to equation of circle C2,

{(x - 3)^2} + {(y - 2)^2} = 5

\Rightarrow {(x - 3)^2} + {x^2} = 5

\Rightarrow {x^2} - 6x + 9 + {x^2} = 5

\Rightarrow 2{x^2} - 6x + 4 = 0

\Rightarrow {x^2} - 3x + 2 = 0

\Rightarrow (x - 2)(x - 1) = 0

\Rightarrow x = 1,2

when

x = 1,

y = 3

and when

x = 2,

y = 4

$\therefore$ Point

A(1,3)

and

B(2,4)

Now, equation of tangent at

A(1,3)

on circle

{(x - 3)^2} + {(y - 2)^2} = 5

or

{x^2} + {y^2} - 6x - 4y + 8 = 0

is

T = 0

x{x_1} + y{y_1} + g(x + {x_1}) + f(y + {y_1}) + C = 0

\Rightarrow x + 3y - 3(x + 1) - 2(y + 3) + 8 = 0

\Rightarrow x + 3y - 3x - 3 - 2y - 6 + 8 = 0

\Rightarrow - 2x + y - 1 = 0

\Rightarrow 2x - y + 1 = 0

....... (1) Similarly tangent at

B(2,4)

is

2x + 4y - 3(x + 2) - 2(y + 4) + 8 = 0

\Rightarrow 2x + 4y - 3x - 6 - 2y - 8 + 8 = 0

\Rightarrow - x + 2y - 6 = 0

\Rightarrow x - 2y + 6 = 0

...... (2) Solving equation (1) and (2), we get

x - 2(2x + 1) + 6 = 0

\Rightarrow x - 4x - 2 + 6 = 0

\Rightarrow - 3x + 4 = 0

\Rightarrow x = {4 \over 3}

$\therefore$

y = 2 \times {4 \over 3} + 1 = {{11} \over 3}

$\therefore$ Point

N = \left( {{4 \over 3},{{11} \over 3}} \right)

Now area of the triangle ANB

= {1 \over 2}\left| \begin{array}{lll}{{x_1}} & {{y_1}} & 1 \\ {{x_2}} & {{y_2}} & 1 \\ {{x_3}} & {{y_3}} & 1 \end{array} \right|

= {1 \over 2}\left| \begin{array}{lll}1 & 3 & 1 \\ 2 & 4 & 1 \\ {{4 \over 3}} & {{{11} \over 3}} & 1 \end{array} \right|

= {1 \over 2}\left[ {1\left( {4 - {{11} \over 3}} \right) - 3\left( {2 - {4 \over 3}} \right) + 1\left( {{{22} \over 3} - {{16} \over 3}} \right)} \right]

= {1 \over 2}\left[ {{1 \over 3} - {6 \over 3} + {6 \over 3}} \right]

= {1 \over 6}

Q75

If the circle

x^{2}+y^{2}-2 g x+6 y-19 c=0, g, c \in \mathbb{R}

passes through the point

(6,1)

and its centre lies on the line

x-2 c y=8

, then the length of intercept made by the circle on

x

-axis is :

A

\sqrt{11}

B 4

C 3

D

2 \sqrt{23}

Correct Answer

Option D

Solution

Circle :

{x^2} + {y^2} - 2gx + 6y - 19c = 0

It passes through

h(6,1)

\Rightarrow 36 + 1 - 12g + 6 - 19c = 0

= 12g + 19c = 43

..... (1) Line

x - 2cy = 8

passes through centre

\Rightarrow g + 6c = 8

...... (2) From (1) & (2)

g = 2,\,c = 1

C:{x^2} + {y^2} - 4x + 6y - 19 = 0

x intercept

= 2\sqrt {{g^2} - C}

= 2\sqrt {4 + 19}

= 2\sqrt {23}

Q76

Let a triangle ABC be inscribed in the circle

{x^2} - \sqrt 2 (x + y) + {y^2} = 0

such that

\angle BAC = {\pi \over 2}

. If the length of side AB is

\sqrt 2

, then the area of the

\Delta

ABC is equal to :

A 1

B

\left( {\sqrt 6 + \sqrt 3 } \right)/2

C

\left( {3 + \sqrt 3 } \right)/4

D

\left( {\sqrt 6 + 2\sqrt 3 } \right)/4

Correct Answer

Option A

Solution

Note: For equation of circle

{x^2} + {y^2} + 2gx + 2fy + c = 0

, center is

( - g,\, - f)

and radius

r = \sqrt {{g^2} + {f^2} - c}

Given, equation of circle is

{x^2} - \sqrt 2 (x + y) + {y^2} = 0

\Rightarrow {x^2} + {y^2} - \sqrt 2 x - \sqrt 2 y = 0

\Rightarrow {x^2} + {y^2} + 2\left( { - {1 \over {\sqrt 2 }}} \right)x + 2\left( { - {1 \over {\sqrt 2 }}} \right) = 0

$\therefore$

g = - {1 \over {\sqrt 2 }}

and

f = - {1 \over {\sqrt 2 }}

$\therefore$ Center

= ( - g,\, - f) = \left( {{1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }}} \right)

And Radius

= r = \sqrt {{{\left( { - {1 \over {\sqrt 2 }}} \right)}^2} + {{\left( { - {1 \over {\sqrt 2 }}} \right)}^2} - 0}

= \sqrt {{1 \over 2} + {1 \over 2}} = \sqrt 1 = 1

As AB and AC makes an angle 90

^\circ

then line BC passes through the center of circle and BC is the diameter of the circle.

$\therefore$ Length of BC = 2r = 2 $\times$ 1 = 2 $\therefore$ AC2 = BC2 $-$ AB2 = 22 $-$ (

\sqrt2

)2 = 2 $\Rightarrow$ AC =

\sqrt2

$\therefore$ Area of right angle triangle ABC =

{1 \over 2}

$\times$ AC $\times$ AB =

{1 \over 2}

$\times$

\sqrt2

\sqrt2

= 1 square unit.

Q77

If the tangents drawn at the points

O(0,0)

and

P\left( {1 + \sqrt 5 ,2} \right)

on the circle

{x^2} + {y^2} - 2x - 4y = 0

intersect at the point Q, then the area of the triangle OPQ is equal to :

A

{{3 + \sqrt 5 } \over 2}

B

{{4 + 2\sqrt 5 } \over 2}

C

{{5 + 3\sqrt 5 } \over 2}

D

{{7 + 3\sqrt 5 } \over 2}

Correct Answer

Option C

Solution

\tan 2\theta = 2 \Rightarrow {{2\tan \theta } \over {1 - {{\tan }^2}\theta }} = 2

\tan \theta = {{\sqrt 5 - 1} \over 2}

(as $\theta$ is acute) Area

= {1 \over 2}{L^2}\sin 2\theta = {1 \over 2}.\,{5 \over {{{\tan }^2}\theta }}.\,2\sin \theta \cos \theta

= {{5\sin \theta \cos \theta } \over {{{\sin }^2}\theta }}.\,{\cos ^2}\theta

= 5\cot \theta .\,{\cos ^2}\theta

= 5.\,{2 \over {\sqrt 5 - 1}}.\,{1 \over {1 + {{\left( {{{\sqrt 5 - 1} \over 2}} \right)}^2}}}

= {{10} \over {\sqrt 5 - 1}}.\,{4 \over {4 + 6 - 2\sqrt 5 }}

= {{40} \over {2\sqrt 5 {{(\sqrt 5 - 1)}^2}}} = {{4\sqrt 5 } \over {6 - 2\sqrt 5 }}

= {{4\sqrt 5 (6 + 2\sqrt 5 )} \over {16}}

= {{\sqrt 5 (3 + \sqrt 5 )} \over 2}

Q78

The set of values of k, for which the circle

C:4{x^2} + 4{y^2} - 12x + 8y + k = 0

lies inside the fourth quadrant and the point

\left( {1, - {1 \over 3}} \right)

lies on or inside the circle C, is :

A an empty set

B

\left( {6,{{65} \over 9}} \right]

C

\left[ {{{80} \over 9},10} \right)

D

\left( {9,{{92} \over 9}} \right]

Correct Answer

Option D

Solution

C:4{x^2} + 4{y^2} - 12x + 8y + k = 0

$\because$

\left( {1, - {1 \over 3}} \right)

lies on or inside the C then

4 + {4 \over 9} - 12 - {8 \over 3} + k \le 0

\Rightarrow k \le {{92} \over 9}

Now, circle lies in 4th quadrant centre

\equiv \left( {{3 \over 2}, - 1} \right)

$\therefore$

r

\Rightarrow {{13} \over 4} - {k \over 4}

\Rightarrow {k \over 4} > {9 \over 4}

\Rightarrow k > 9

$\therefore$

k \in \left( {9,{{92} \over 9}} \right)

Q79

Let C be a circle passing through the points A(2,

-

1) and B(3, 4). The line segment AB s not a diameter of C. If r is the radius of C and its centre lies on the circle

{(x - 5)^2} + {(y - 1)^2} = {{13} \over 2}

, then r2 is equal to :

A 32

B

{{65} \over 2}

C

{{61} \over 2}

D 30

Correct Answer

Option B

Solution

Equation of perpendicular bisector of AB is

y - {3 \over 2} = - {1 \over 5}\left( {x - {5 \over 2}} \right) \Rightarrow x + 5y = 10

Solving it with equation of given circle,

{(x - 5)^2}{\left( {{{10 - x} \over 5} - 1} \right)^2} = {{13} \over 2}

\Rightarrow {(x - 5)^2}\left( {1 + {1 \over {25}}} \right) = {{13} \over 2}

\Rightarrow x - 5 = \pm \,{5 \over 2} \Rightarrow x = {5 \over 2}

or

{{15} \over 2}

But

x \ne {5 \over 2}

because AB is not the diameter. So, centre will be

\left( {{{15} \over 2},{1 \over 2}} \right)

Now,

{r^2} = {\left( {{{15} \over 2} - 2} \right)^2} + {\left( {{1 \over 2} + 1} \right)^2}

= {{65} \over 2}

Q80

A circle touches both the y-axis and the line x + y = 0. Then the locus of its center is :

A

y = \sqrt 2 x

B

x = \sqrt 2 y

C

{y^2} - {x^2} = 2xy

D

{x^2} - {y^2} = 2xy

Correct Answer

Option D

Solution

Let the centre be (h, k) So,

\left| h \right| = \left| {{{h + k} \over {\sqrt 2 }}} \right|

\Rightarrow 2{h^2} = {h^2} + {k^2} + 2hk

Locus will be

{x^2} - {y^2} = 2xy