Circle Questions — JEE Mathematics

Q1

Intercept on the line y = x by the circle

{x^2}\, + \,{y^2} - 2x = 0

is AB. Equation of the circle on AB as a diameter is :

A

\,{x^2}\, + \,{y^2} + \,x\, - \,y\,\, = 0

B

\,{x^2}\, + \,{y^2} - \,x\, + \,y\,\, = 0

C

\,{x^2}\, + \,{y^2} + \,x\, + \,y\,\, = 0

D

\,{x^2}\, + \,{y^2} - \,x\, - \,y\,\, = 0

Correct Answer

Option D

Solution

Solving

y=x

and the circle

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

we get

x = 0,y = 0

and

x=1,

y=1

$\therefore$ Extremities of diameter of the required circle are

\left( {0,0} \right)

and

\left( {1,1} \right)

. Hence, the equation of circle is

\left( {x - 0} \right)\left( {x - 1} \right) + \left( {y - 0} \right)\left( {y - 1} \right) = 0

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

Q2

If the chord y = mx + 1 of the circle

{x^2}\, + \,{y^2} = 1

subtends an angle of measure

{45^ \circ }

at the major segment of the circle then value of m is :

A

2\, \pm \,\sqrt 2 \,\,

B

- \,2\, \pm \,\sqrt 2 \,

C

- 1\, \pm \,\sqrt 2 \,\,

D none of these

Correct Answer

Option C

Solution

Equation of circle

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

\Rightarrow {x^2} + {y^2} = {\left( {y - mx} \right)^2}

\Rightarrow {x^2} = {m^2}{x^2} - 2\,\,mxy;

\Rightarrow {x^2}\left( {1 - {m^2}} \right) + 2mxy = 0.

Which represents the pair of lines between which the angle is

{45^ \circ }.

\tan 45 = \pm {{2\sqrt {{m^2} - 0} } \over {1 - {m^2}}} = {{ \pm 2m} \over {1 - {m^2}}};

\Rightarrow 1 - {m^2} = \pm 2m

\Rightarrow {m^2} \pm 2m - 1 = 0

\Rightarrow m = {{ - 2 \pm \sqrt {4 + 4} } \over 2}

= {{ - 2 \pm 2\sqrt 2 } \over 2}

= - 1 \pm \sqrt 2 .

Q3

The two adjacent sides of a cyclic quadrilateral are 2 and 5 and the angle between them is 60o. If the area of the quadrilateral is

4\sqrt 3

, then the perimeter of the quadrilateral is :

A 12.5

B 13.2

C 12

D 13

Correct Answer

Option C

Solution

Here; cos $\theta$ =

{{{a^2} + {b^2} - {c^2}} \over {2ab}}

and $\theta$ = 60o $\Rightarrow$ cos 60o =

{{4 + 25 - {c^2}} \over {2.2.5}}

$\Rightarrow$ 10 = 29 $-$ c2 $\Rightarrow$ c2 = 19 $\Rightarrow$ c =

\sqrt {19}

also; cos $\theta$ =

{{{a^2} + {b^2} - {c^2}} \over {2ab}}

and $\theta$ = 120o $\Rightarrow$ $-$

{1 \over 2}

=

{{{a^2} + {b^2} - 19} \over {2ab}}

$\Rightarrow$ a2 + b2 $-$ 19 = $-$ ab $\Rightarrow$ a2 + b2 + ab = 19 $\therefore$ Area =

{1 \over 2} \times 2 \times 5

sin 60 +

{1 \over 2}

ab sin 120o = 4

\sqrt 3

$\Rightarrow$

{{5\sqrt 3 } \over 2} + {{ab\sqrt 3 } \over 4}

=

4\sqrt 3

$\Rightarrow$

{{ab} \over 4}

= 4 $-$

{5 \over 2}

=

{3 \over 2}

$\Rightarrow$ ab = 6 $\therefore$ a2 + b2 = 13 $\Rightarrow$ a = 2, b = 3 Perimeter = Sum of all sides = 2 + 5 + 2 + 3 = 12

Q4

A line drawn through the point P(4, 7) cuts the circle x2 + y2 = 9 at the points A and B. Then PA⋅PB is equal to :

A 53

B 56

C 74

D 65

Correct Answer

Option B

Solution

P(4, 7). Here, x = 4, y = 7 $\therefore$ PA $\times$ PB = PT2 Also; PT =

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

$\Rightarrow$ PT =

\sqrt {16 + 49 - 9}

=

\sqrt {56}

$\Rightarrow$ PT2 = 56 $\therefore$ PA $\times$ PB = 56

Q5

If one of the diameters of the circle, given by the equation,

{x^2} + {y^2} - 4x + 6y - 12 = 0,

is a chord of a circle

S

, whose centre is at

(-3, 2)

, then the radius of

S

is :

A

5

B

10

C

5\sqrt 2

D

5\sqrt 3

Correct Answer

Option D

Solution

Center of

S

:

O(-3, 2)

center of given circle

A(2, -3)

\Rightarrow OA = 5\sqrt 2

Also

AB=5

(as

AB=r

of the given circle) $\Rightarrow$ Using pythagoras theorem in

\Delta OAB

r = 5\sqrt 3

Q6

Equation of the tangent to the circle, at the point (1, −1), whose centre is the point of intersection of the straight lines x − y = 1 and 2x + y = 3 is :

A 4x + y − 3 = 0

B x + 4y + 3 = 0

C 3x − y − 4 = 0

D x − 3y − 4 = 0

Correct Answer

Option B

Solution

Point of intersection of lines x $-$ y = 1 and 2x + y = 3 is

\left( {{4 \over 3},{1 \over 3}} \right)

Slope of OP =

{{{1 \over 3} + 1} \over {{4 \over 3} - 1}}

=

{{{4 \over 3}} \over {{1 \over 3}}}

= 4 Slope of tangent = $-$

{1 \over 4}

Equation of tangent y + 1 = $-$

{1 \over 4}

(x $-$ 1) 4y + 4 = $-$ x + 1 x + 4y + 3 = 0

Q7

The number of common tangents to the circles

{x^2} + {y^2} - 4x - 6x - 12 = 0

and

{x^2} + {y^2} + 6x + 18y + 26 = 0,

is :

A

3

B

4

C

1

D

2

Correct Answer

Option A

Solution

{x^2} + {y^2} - 4x - 6y - 12 = 0\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)

Center,

{c_1} = \left( {2,\,3} \right)

and Radius,

{r_1} = 5

units

{x^2} + {y^2} + 6x + 18y + 26 = 0\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

Center,

{c_2} = \left( { - 3, - 9} \right)

and Radius,

{r_2} = 8

units

{C_1}{C_2} = \sqrt {{{\left( {2 + 3} \right)}^2} + {{\left( {3 + 9} \right)}^2}} = 13\,\,

units

{r_1} + {r_2} = 5 + 8 = 13

$\therefore$

{C_1}{C_2} = {r_1} + {r_2}

Therefore there are three common tangents.

Q8

The centre of the circle passing through (0, 0) and (1, 0) and touching the circle

{x^2}\, + \,{y^2} = 9

is :

A

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

B

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

C

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

D

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

Correct Answer

Option B

Solution

Let the required circle be

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

Since it passes through

\left( {0,0} \right)

and

\left( {1,0} \right)

\Rightarrow c = 0

and

g = - {1 \over 2}

Points

\left( {0,0} \right)

and

\left( {1,0} \right)

lie inside the circle

{x^2} + {y^2} = 9,

so two circles touch internally

\Rightarrow c{}_1{c_2} = {r_1} - {r_2}

$\therefore$

\sqrt {{g^2} + {f^2}} = 3 - \sqrt {{g^2} + {f^2}}

\Rightarrow \sqrt {{g^2} + {f^2}} = {3 \over 2}

\Rightarrow {f^2} = {9 \over 4} - {1 \over 4} = 2

$\therefore$

f = \pm \sqrt 2 .

Hence, the centers of required circle are

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

or

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

Q9

The centres of those circles which touch the circle,

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

, externally and also touch the

x

-axis, lie on :

A a circle

B an ellipse which is not a circle

C a hyperbola

D a parabola

Correct Answer

Option D

Solution

For the given circle, center :

(4,4)

radius

=6

6 + k = \sqrt {{{\left( {h - 4} \right)}^2} + {{\left( {k - 4} \right)}^2}}

{\left( {h - 4} \right)^2} = 20k + 20

$\therefore$ locus of

(h, k)

is

{\left( {h - 4} \right)^2} = 20\left( {y + 1} \right),

which is parabola.

Q10

The equation of a circle with origin as a center and passing through an equilateral triangle whose median is of length

3

a

is :

A

{x^2}\, + \,{y^2} = 9{a^2}

B

{x^2}\, + \,{y^2} = 16{a^2}

C

{x^2}\, + \,{y^2} = 4{a^2}

D

{x^2}\, + \,{y^2} = {a^2}

Correct Answer

Option C

Solution

Let

ABC

be an equilateral triangle, whose median is

AD.

Given

AD=3a.

In

\Delta ABD,\,\,A{B^2} = A{D^2} + B{D^2};

\Rightarrow {x^2} = 9{a^2} + \left( {{x^2}/4} \right)\,\,

where

AB = BC = AC = x.

{3 \over 4}{x^2} = 9{a^2} \Rightarrow {x^2} = 12{a^2}.

In

\,\,\,\Delta OBD,O{B^2} = O{D^2} + B{D^2}

\Rightarrow {r^2} = {\left( {3a - r} \right)^2} + {{{x^2}} \over 4}

\Rightarrow {r^2} = 9{a^2} - 6ar + {r^2} + 3{a^2};

\Rightarrow 6ar = 12{a^2}

\Rightarrow r = 2a

So equation of circle is

{x^2} + {y^2} = 4{a^2}