Circle — Page 4 | JEE Mathematics

Q31

The circle passing through

(1, -2)

and touching the axis of

x

at

(3, 0)

also passes through the point :

A

\left( { - 5,\,2} \right)

B

\left( { 2,\,-5} \right)

C

\left( { 5,\,-2} \right)

D

\left( { - 2,\,5} \right)

Correct Answer

Option C

Solution

Since circle touches

x

-axis at

(3,0)

$\therefore$ The equation of circle be

{\left( {x - 3} \right)^2} + {\left( {y - 0} \right)^2} + \lambda y = 0

As it passes through

(1, -2)

$\therefore$ Put

x=1,

y=-2

\Rightarrow {\left( {1 - 3} \right)^2} + {\left( { - 2} \right)^2} + \lambda \left( { - 2} \right) = 0

\Rightarrow \lambda = 4

$\therefore$ equation of circle is

{\left( {x - 3} \right)^2} + {y^2} - 8 = 0

Now, from the options

\left( {5, - 2} \right)

satisfies equation of circle.

Q32

Let A(1, 4) and B(1,

-

5) be two points. Let P be a point on the circle (x

-

1)2 + (y

-

1)2 = 1 such that (PA)2 + (PB)2 have maximum value, then the points, P, A and B lie on :

A a straight line

B an ellipse

C a parabola

D a hyperbola

Correct Answer

Option A

Solution

P be a point on

{(x - 1)^2} + {(y - 1)^2} = 1

so

P(1 + \cos \theta ,1 + \sin \theta )

A(1, 4), B(1, $-$ 5)

{(PA)^2} + {(PB)^2}

= {(\cos \theta )^2} + {(\sin \theta - 3)^2} + {(\cos \theta )^2} + {(\sin \theta + 6)^2}

= 47 + 6\sin \theta

It is maximum if

\sin \theta = 1

When

\sin \theta = 1,\cos \theta = 0

So P(1, 2), A(1, 4), B(1, $-$ 5) P, A, B are collinear points.

Q33

Let

C

be the circle with centre at

(1, 1)

and radius

=

1

. If

T

is the circle centred at

(0, y)

, passing through origin and touching the circle

C

externally, then the radius of

T

is equal to :

A

{1 \over 2}

B

{1 \over 4}

C

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

D

{{\sqrt 3 } \over 2}

Correct Answer

Option B

Solution

Equation of circle

C \equiv {\left( {x - 1} \right)^2} + {\left( {y - 1} \right)^2} = 1

Radius of

T = \left| y \right|

T

touches

C

externally therefore, Distance between the centers $=$ sum of their radii

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

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

\Rightarrow 1 + {y^2} + 1 - 2y = 1 + {y^2} + 2\left| y \right|

2\left| y \right| = 1 - 2y

If

y>0

then

2y=1-2y

\Rightarrow y = {1 \over 4}

y<0

then

-2y=1-2y

\Rightarrow 0 = 1

(not possible) $\therefore$

y = {1 \over 4}

Q34

Locus of the image of the point

(2, 3)

in the line

\left( {2x - 3y + 4} \right) + k\left( {x - 2y + 3} \right) = 0,\,k \in R,

is a :

A circle of radius

\sqrt 2

.

B circle of radius

\sqrt 3

.

C straight line parallel to

x

-axis

D straight line parallel to

y

-axis

Correct Answer

Option A

Solution

Intersection point of

2x - 3y + 4 = 0

and

x-2y+3=0

is

(1, 2)

Since,

P

is the fixed point for given family of lines So,

PB=PA

{\left( {\alpha - 1} \right)^2} + {\left( {\beta - 2} \right)^2} = {\left( {2 - 1} \right)^2} + {\left( {3 - 2} \right)^2}

{\left( {\alpha - 1} \right)^2} + {\left( {\beta - 2} \right)^2} = 1 + 1 = 2

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

{\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}

Therefore, given locus is a circle with center

(1, 2)

and radius

\sqrt 2 .

Q35

The radius of a circle, having minimum area, which touches the curve y = 4 – x2 and the lines, y = |x| is :

A

2\left( {\sqrt 2 - 1} \right)

B

4\left( {\sqrt 2 - 1} \right)

C

4\left( {\sqrt 2 + 1} \right)

D

2\left( {\sqrt 2 + 1} \right)

Correct Answer

Option B

Solution

Let the radius of circle with least area be r.

Then, the coordinate of the center = (0, b) $\therefore$ The equation of circle be x2 + (y – b)2 = r2 Distance of perpendiculur from (0, 4) to y = x line = r $\Rightarrow$

\left| {{{ - b} \over {\sqrt 2 }}} \right| = r

$\Rightarrow$ b =

{\sqrt 2 r}

Circle passes through (0, 4), $\therefore$ 0 + (4 – b)2 = r2 $\Rightarrow$ 4 - b = r $\Rightarrow$ 4 -

{\sqrt 2 r}

= r $\Rightarrow$ r =

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

=

4\left( {\sqrt 2 - 1} \right)

Q36

If a circle C, whose radius is 3, touches externally the circle,

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

at the point (2, 2), then the length of the intercept cut by this circle C, on the x-axis is equal to :

A

2\sqrt 5

B

3\sqrt 2

C

\sqrt 5

D

2\sqrt 3

Correct Answer

Option A

Solution

Given circle is : x2 + y2 + 2x $-$ 4y $-$ 4 = 0

\therefore\,\,\,

its center is ( $-$ 1, 2) and radius is 3 units. Let A = (x, y) be the center of the circle C $\therefore$

\,\,\,

{{x - 1} \over 2}

= 2 $\Rightarrow$ x = 5 and

{{y + 2} \over 2}

= 2 $\Rightarrow$ y = 2 So the center of C is (5, 2) and its radius is 3

\therefore\,\,\,

Equation of center C is : x2 + y2 $-$ 10x $-$ 4y + 20 = 0

\therefore\,\,\,

The length of the intercept it cuts on the x-axis = 2

\sqrt {{g^2} - c} = 2\sqrt {25 - 20} = 2\sqrt 5

Q37

For the four circles M, N, O and P, following four equations are given : Circle M : x2 + y2 = 1 Circle N : x2 + y2

-

2x = 0 Circle O : x2 + y2

-

2x

-

2y + 1 = 0 Circle P : x2 + y2

-

2y = 0 If the centre of circle M is joined with centre of the circle N, further center of circle N is joined with centre of the circle O, centre of circle O is joined with the centre of circle P and lastly, centre of circle P is joined with centre of circle M, then these lines form the sides of a :

A Rhombus

B Square

C Rectangle

D Parallelogram

Correct Answer

Option B

Solution

{C_M} = (0,0)

{C_N} = (1,0)

{C_O} = (1,1)

{C_P} = (0,1)

Q38

A circle passes through the points (2, 3) and (4, 5). If its centre lies on the line,

y - 4x + 3 = 0,

then its radius is equal to :

A 2

B

\sqrt 5

C

\sqrt 2

D 1

Correct Answer

Option A

Solution

Centre $(\alpha, \beta)$ lies on line $y-4 x+3=0$ Then, $\quad \beta=4 \alpha-3$ And

\text{ Radius }=\sqrt{(\alpha-2)^2+(\beta-3)^2}=\sqrt{(\alpha-4)^2+(\beta-5)^2}

\begin{aligned} & \alpha^2+\beta^2+ 13-4 \alpha-6 \beta=\alpha^2+\beta^2+41-8 \alpha-10 \beta \\\\ & 4 \alpha+4 \beta=28 \Rightarrow \alpha+\beta=7 \\\\ & \Rightarrow \alpha+4 \alpha-3=7 \\\\ & \Rightarrow \alpha=2, \beta=5 \end{aligned}

Therefore, $\quad$ Radius $=\sqrt{(2-2)^2+(5-3)^2}=2$

Q39

If the tangent at (1, 7) to the curve x2 = y - 6 touches the circle x2 + y2 + 16x + 12y + c = 0, then the value of c is :

A 95

B 195

C 185

D 85

Correct Answer

Option A

Solution

NOTE : Equation of tangent at (x1, y1) to the curve x2 = 4ay is xx1 = 4a

\left( {{{y + {y_1}} \over 2}} \right)

Now equation of tangent at (1, 7) to x2 = y $-$ 6 is

x\,.\,1 = 4\,.\,{1 \over 4}\left( {{{y + 7} \over 2}} \right) - 6

\Rightarrow 2x = y + 7 - 12

\Rightarrow 2x - y + 5 = 0

This tangent touches the circle.

So, perpendicular distance from the center of the circle to the tangent is equal to the radius of the circle.

For the circle,

{x^2} + {y^2} + 16x + 12y + C = 0

center is ( $-$ 8, $-$ 6) and radius (r) =

\sqrt {{8^2} + {6^2} - c}

= \sqrt {100 - c}

Distance of the tangent from the center of the circle d =

\left| {{{2\left( { - 8} \right) - \left( { - 6} \right) + 5} \over {\sqrt {{2^2} + {1^2}} }}} \right|

= \left| {{{ - 16 + 6 + 5} \over {\sqrt 5 }}} \right|

And we know d = r

\therefore\,\,\,

\left| {{{ - 16 + 11} \over {\sqrt 5 }}} \right| = \sqrt {100 - c}

\Rightarrow \left| {{{ - 5} \over {\sqrt 5 }}} \right| = \sqrt {100 - c}

\Rightarrow \left| { - \sqrt 5 } \right| = \sqrt {100 - c}

\Rightarrow 5 = 100 - c

\Rightarrow c = 95

Q40

If the locus of the mid-point of the line segment from the point (3, 2) to a point on the circle, x2 + y2 = 1 is a circle of radius r, then r is equal to :

A

{1 \over 4}

B

{1 \over 2}

C 1

D

{1 \over 3}

Correct Answer

Option B

Solution

Let P(h, k) and point on the circle is (cos $\theta$ , sin $\theta$ ) $\therefore$

{{3 + \cos \theta } \over 2} = h

and

{{2 + \sin \theta } \over 2} = k

cos $\theta$ = 2h $-$ 3 and sin $\theta$ = 2h $-$ 2 Squaring and adding we get

{(2h - 3)^2} + {(2h - 2)^2} = 1

\Rightarrow 4{x^2} - 12x + 9 + 4{y^2} - 8y + 4 = 1

\Rightarrow 4{x^2} + 4{y^2} - 12x - 8y + 12 = 0

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

Radius =

\sqrt {{9 \over 4} + 1 - 3} = {1 \over 2}