Circle — Page 6 | JEE Mathematics

Q51

The tangent and the normal lines at the point (

\sqrt 3

, 1) to the circle x2 + y2 = 4 and the x-axis form a triangle. The area of this triangle (in square units) is :

A

{4 \over {\sqrt 3 }}

B

{1 \over {\sqrt 3 }}

C

{2 \over {\sqrt 3 }}

D

{1 \over {3 }}

Correct Answer

Option C

Solution

Equation of tangent to the circle x2 + y2 = 4 at point (

\sqrt 3

, 1) is

\sqrt 3

x + y = 4 Slope of this tangent (m) = -

\sqrt 3

$\therefore$ Slope of the normal at point (

\sqrt 3

, 1) is =

{1 \over {\sqrt 3 }}

$\therefore$ Equation of normal at point (

\sqrt 3

, 1), y - 1 =

{1 \over {\sqrt 3 }}\left( {x - \sqrt 3 } \right)

$\Rightarrow$ x -

\sqrt 3

y = 0 So normal passes through the center of the axis.

Tangent cut the x-axis at point A when y = 0, So the x coordinate of the point A is

\sqrt 3

x + 0 = 4 $\Rightarrow$ x =

{4 \over {\sqrt 3 }}

$\therefore$ Coordinate of A =

\left( {{4 \over {\sqrt 3 }},0} \right)

Area of triangle OAB,

\Delta

=

{1 \over 2}\left| \begin{array}{lll}0 & 0 & 1 \\ {{4 \over {\sqrt 3 }}} & 0 & 1 \\ {\sqrt 3 } & 1 & 1 \end{array} \right|

=

{1 \over 2}\left( {{4 \over {\sqrt 3 }}} \right)

=

{{2 \over {\sqrt 3 }}}

Q52

The sum of the squares of the lengths of the chords intercepted on the circle, x2 + y2 = 16, by the lines, x + y = n, n

\in

N, where N is the set of all natural numbers, is :

A 210

B 160

C 320

D 105

Correct Answer

Option A

Solution

Let the chord x + y = n cuts the circle x2 + y2 = 16 at A and B. Length of perpendicular from O on AB, OM =

\left| {{{0 + 0 - n} \over {\sqrt {{1^2} + {1^2}} }}} \right| = {n \over {\sqrt 2 }}

Radius of the circle = 4 From the picture you can see,

{n \over {\sqrt 2 }}

< 4 $\Rightarrow$ n < 5.65 $\therefore$ Possible value of n = 1, 2, 3, 4, 5 Length of chord AB =

2\sqrt {{{\left( 4 \right)}^2} - {{\left( {{n \over {\sqrt 2 }}} \right)}^2}}

=

2\sqrt {16 - {{{n^2}} \over 2}}

=

\sqrt {64 - 2{n^2}}

=

{l}

For n = 1,

{l^2}

= 62 For n = 2,

{l^2}

= 56 For n = 3,

{l^2}

= 46 For n = 4,

{l^2}

= 32 For n = 5,

{l^2}

= 14 $\therefore$ Sum of square of length of chords = 62 + 56 + 46 + 32 + 14 = 210

Q53

A square is inscribed in the circle x2 + y2 – 6x + 8y – 103 = 0 with its sides parallel to the coordinate axes. Then the distance of the vertex of this square which is nearest to the origin is :

A

\sqrt {137}

B 6

C

\sqrt {41}

D 13

Correct Answer

Option C

Solution

R

= \sqrt {9 + 16 + 103} = 8\sqrt 2

OA

= 13

OB

= \sqrt {265}

OC

= \sqrt {137}

OD

= \sqrt {41}

Q54

If the length of the chord of the circle, x2 + y2 = r2 (r > 0) along the line, y – 2x = 3 is r, then r2 is equal to :

A

{9 \over 5}

B

{{24} \over 5}

C

{{12} \over 5}

D 12

Correct Answer

Option C

Solution

In right

\Delta

CDB, sin 60o =

{{CD} \over r}

$\Rightarrow$ CD =

r \times {{\sqrt 3 } \over 2}

=

{{\sqrt 3 r} \over 2}

Now equation of AB is y – 2x – 3 = 0 So

{{\sqrt 3 r} \over 2}

=

{{\left| {0 + 0 - 3} \right|} \over {\sqrt 5 }}

$\Rightarrow$

{{\sqrt 3 r} \over 2}

=

{3 \over {\sqrt 5 }}

$\Rightarrow$ r =

{{2\sqrt 3 } \over 5}

$\Rightarrow$ r2 =

{{12} \over 5}

Q55

The circle passing through the intersection of the circles, x2 + y2 – 6x = 0 and x2 + y2 – 4y = 0, having its centre on the line, 2x – 3y + 12 = 0, also passes through the point :

A (–3, 1)

B (1, –3)

C (–1, 3)

D (–3, 6)

Correct Answer

Option D

Solution

Let S be the circle passing through point of intersection of S1 & S2 $\therefore$ S = S1 + $\lambda$ S2 = 0 $\Rightarrow$

S:({x^2} + {y^2} - 6x) + \lambda ({x^2} + {y^2} - 4y) = 0

$\Rightarrow$

S:{x^2} + {y^2} - \left( {{6 \over {1 + \lambda }}} \right)x - \left( {{{4\lambda } \over {1 + \lambda }}} \right)y = 0

....(1) Centre

\left( {{3 \over {1 + \lambda }},{{2\lambda } \over {1 + \lambda }}} \right)

lies on

2x - 3y + 12 = 0 \Rightarrow \lambda = - 3

put in

(1) \Rightarrow S:{x^2} + {y^2} + 3x - 6y = 0

Now check options point

( - 3,6)

lies on S.

Q56

A circle touches the y-axis at the point (0, 4) and passes through the point (2, 0). Which of the following lines is not a tangent to this circle?

A 3x – 4y – 24 = 0

B 4x + 3y – 8 = 0

C 3x + 4y – 6 = 0

D 4x – 3y + 17 = 0

Correct Answer

Option B

Solution

Equation of family of circle touching y-axis at (0, 4) is given by (x – 0)2 + (y – 4)2 + $\lambda$ x = 0.

$\because$ It passes through (2, 0) $\Rightarrow$ $\lambda$ = –10.

$\Rightarrow$ Required circle is (x – 0)2 + (y – 4)2 – 10x = 0 $\Rightarrow$ x2 + y2 – 10x – 8y + 16 = 0 $\therefore$ center of circle (5, 4) and radius = 5 By checking all options you can see 4x + 3y – 8 = 0 is not a tangent to the circle.

As distance of 4x + 3y – 8 = 0 from (5, 4) =

\left| {{{24} \over 5}} \right|

$\ne$ radius

Q57

If a line, y = mx + c is a tangent to the circle, (x – 3)2 + y2 = 1 and it is perpendicular to a line L1, where L1 is the tangent to the circle, x2 + y2 = 1 at the point

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

, then :

A c2 + 6c + 7 = 0

B c2 - 7c + 6 = 0

C c2 – 6c + 7 = 0

D c2 + 7c + 6 = 0

Correct Answer

Option A

Solution

For circle x2 + y2 = 1 tangnet at point P

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

is T = 0 $\Rightarrow$

{1 \over {\sqrt 2 }}x + {1 \over {\sqrt 2 }}y - 1 = 0

$\Rightarrow$ x + y -

\sqrt 2

= 0 Perpendicular to the line is x - y + c = 0 This is tangent to the circle (x – 3)2 + y2 = 1 Also we know, perpendicular distance from center = radius $\therefore$

\left| {{{3 - 0 + c} \over {\sqrt 2 }}} \right|

= 1 $\Rightarrow$ |3 + c| =

{\sqrt 2 }

$\Rightarrow$

{\left( {3 + c} \right)^2} = 2

$\Rightarrow$ 9 + c2 + 6c = 2 $\Rightarrow$ c2 + 6c + 7 = 0

Q58

Let the tangents drawn from the origin to the circle, x2 + y2 - 8x - 4y + 16 = 0 touch it at the points A and B. The (AB)2 is equal to :

A

{{56} \over 5}

B

{{32} \over 5}

C

{{52} \over 5}

D

{{64} \over 5}

Correct Answer

Option D

Solution

Equation of chord of contact is x.0 + y.0 – 4(x + 0) –2(y + 0) + 16 = 0 $\Rightarrow$ 2x + y – 8 = 0 $\therefore$ Length of CM =

\left[ {{{2.4 + 2 - 8} \over {\sqrt {{2^2} + {1^2}} }}} \right]

=

{2 \over {\sqrt 5 }}

units $\therefore$ AM = BM =

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

=

\sqrt {{{16} \over 5}}

$\therefore$ Length of chord of contact (AB) =

{8 \over {\sqrt 5 }}

$\therefore$ AB2 =

{{16 \times 16} \over {20}}

=

{{64} \over 5}

Q59

Let the lengths of intercepts on x-axis and y-axis made by the circle x2 + y2 + ax + 2ay + c = 0, (a < 0) be 2

{\sqrt 2 }

and 2

{\sqrt 5 }

, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line x + 2y = 0, is equal to :

A

{\sqrt {10} }

B

{\sqrt {6} }

C

{\sqrt {11} }

D

{\sqrt {7} }

Correct Answer

Option B

Solution

2\sqrt {{{{a^2}} \over 4} - c} = 2\sqrt 2

\sqrt {{a^2} - 4c} = 2\sqrt 2

{a^2} - 4c = 8

.... (1)

2\sqrt {{a^2} - c} = 2\sqrt 5

{a^2} - c = 5

.... (2)

(2) - (1)

3c = - 3a \Rightarrow c = - 1

{a^2} = 4 \Rightarrow a = - 2

(Given a < 0) Equation of circle

{x^2} + {y^2} - 2x - 4y - 1 = 0

Equation of tangent which is perpendicular to the line x + 2y = 0 is

2x - y + \lambda = 0

$\therefore$ p = r

\left| {{{2 - 2 + \lambda } \over {\sqrt 5 }}} \right| = \sqrt 6

\Rightarrow \lambda = \pm \sqrt {30}

$\therefore$ Tangent

2x - y \pm \sqrt {30} = 0

Distance from origin =

{{\sqrt {30} } \over {\sqrt 5 }} = \sqrt 6

Q60

The line 2x

-

y + 1 = 0 is a tangent to the circle at the point (2, 5) and the centre of the circle lies on x

-

2y = 4. Then, the radius of the circle is :

A 5

\sqrt 3

B 4

\sqrt 5

C 3

\sqrt 5

D 5

\sqrt 4

Correct Answer

Option C

Solution

{m_1} \times {m_2} = - 1

{{{{a - 4} \over 2} - 5} \over {a - 2}} \times 2 = - 1

{{a - 14} \over {a - 2}} = - 1

a - 14 = 2 - a

2a = 16

a = 8 $\therefore$ Centre (8, 2) Radius =

\sqrt {36 + 9}

= \sqrt {45}

= 3\sqrt 5