Circle — Page 13 | JEE Mathematics

Q121

If two parallel chords of a circle, having diameter 4units, lie on the opposite sides of the center and subtend angles

{\cos ^{ - 1}}\left( {{1 \over 7}} \right)

and sec

-

1 (7) at the center respectivey, then the distance between these chords, is :

A

{4 \over {\sqrt 7 }}

B

{8 \over {\sqrt 7 }}

C

{8 \over 7}

D

{16 \over 7}

Correct Answer

Option B

Solution

Since cos2 $\theta$ = 1/7 $\Rightarrow$ 2 cos2 Q $-$ 1 = 1/7 $\Rightarrow$ 2 cos2 $\theta$ = 8/7 $\Rightarrow$ cos2 $\theta$ = 4/7 $\Rightarrow$ cos2 $\theta$ =

{4 \over 7}

$\Rightarrow$ cos2 $\theta$ =

{2 \over {\sqrt 7 }}

Also, sec2 $\phi$ = 7 =

{1 \over {2{{\cos }^2}\phi - 1}}

7 = cos2 $\phi$ $-$ 1 =

{1 \over 7}

= 2 cos2 $\phi$ =

{8 \over 7}

= cos $\phi$ =

{2 \over {\sqrt 7 }}

P1P2 = r cos $\theta$ + r cos $\phi$ =

{4 \over {\sqrt 7 }} + {4 \over {\sqrt 7 }}

=

{8 \over {\sqrt 7 }}

Q122

The tangent to the circle C1 : x2 + y2

-

2x

-

1 = 0 at the point (2, 1) cuts off a chord of length 4 from a circle C2 whose center is (3,

-

2). The radius of C2 is :

A 2

B

\sqrt 2

C 3

D

\sqrt 6

Correct Answer

Option D

Solution

Here, equation of tangent on C1 at (2, 1) is : 2x + y $-$ (x + 2) $-$ 1 = 0 Or x + y = 3 If it cuts off the chord of the circle C2 then the equation of the chord is : x + y = 3

\therefore\,\,\,

distance of the chord from (3, $-$ 2) is : d =

\left| {{{3 - 2 - 3} \over {\sqrt 2 }}} \right|

=

\sqrt 2

Also, length of the chord is

l

= 4

\therefore\,\,\,

radius of C2 = r =

\sqrt {{{\left( {{l \over 2}} \right)}^2} + {d^2}}

=

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

Q123

If a variable line, 3x + 4y –

\lambda

= 0 is such that the two circles x2 + y2 – 2x – 2y + 1 = 0 and x2 + y2 – 18x – 2y + 78 = 0 are on its opposite sides, then the set of all values of

\lambda

is the interval :

A (23, 31)

B (2, 17)

C [13, 23]

D [12, 21]

Correct Answer

Option D

Solution

Centre of circles are opposite side of line (3 + 4 $-$ $\lambda$ ) (27 + 4 $-$ $\lambda$ ) < 0 ( $\lambda$ $-$ 7) ( $\lambda$ $-$ 31) < 0 $\lambda$

\in

(7, 31) distance from S1

\left| {{{3 + 4 - \lambda } \over 5}} \right| \ge 1 \Rightarrow \lambda \in ( - \infty ,2] \cup [(12,\infty ]

distance from S2

\left| {{{27 + 4 - \lambda } \over 5}} \right| \ge 2 \Rightarrow \lambda \in ( - \infty ,21] \cup [41,\infty )

so

\lambda \in \left[ {12,21} \right]

Q124

Two circles with equal radii are intersecting at the points (0, 1) and (0, –1). The tangent at the point (0, 1) to one of the circles passes through the centre of the other circle. Then the distance between the centres of these circles is :

A

2\sqrt 2

B

\sqrt 2

C 2

D 1

Correct Answer

Option C

Solution

In

\Delta

APO

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

$\Rightarrow$

r = \sqrt 2

So distance between centres

= \sqrt 2 r = 2

Q125

The straight line x + 2y = 1 meets the coordinate axes at A and B. A circle is drawn through A, B and the origin. Then the sum of perpendicular distances from A and B on the tangent to the circle at the origin is :

A

4\sqrt 5

B

{{\sqrt 5 } \over 2}

C

2\sqrt 5

D

{{\sqrt 5 } \over 4}

Correct Answer

Option B

Solution

Equation of circle (x $-$ 1) (x $-$ 0) + (y $-$ 0) (y $-$

{1 \over 2}

) = 0 $\Rightarrow$ x2 + y2 $-$ x $-$

{y \over 2}

= 0 Equation of tangent of region is 2x + y = 0

\ell

1 +

\ell

2 =

{2 \over {\sqrt 5 }} + {1 \over {2\sqrt 5 }}

=

{{4 + 1} \over {2\sqrt 5 }} = {{\sqrt 5 } \over 2}

Q126

If the area of an equilateral triangle inscribed in the circle x2 + y2 + 10x + 12y + c = 0 is

27\sqrt 3

sq units then c is equal to :

A 20

B 25

C

-

25

D 13

Correct Answer

Option B

Solution

3\left( {{1 \over 2}{r^2}.\sin {{120}^o}} \right) = 27\sqrt 3

{{{r^2}} \over 2}{{\sqrt 3 } \over 2} = {{27\sqrt 3 } \over 3}

{r^2} = {{108} \over 3} = 36

Radius

= \sqrt {25 + 36 - C} = \sqrt {36}

C = 25

Q127

If a circle C passing through the point (4, 0) touches the circle x2 + y2 + 4x – 6y = 12 externally at the point (1, – 1), then the radius of C is :

A 5

B 2

\sqrt {5}

C 4

D

\sqrt {37}

Correct Answer

Option A

Solution

x2 + y2 + 4x $-$ 6y $-$ 12 = 0 Equation of tangent at (1, $-$ 1) x $-$ y + 2(x + 1) $-$ 3(y $-$ 1) $-$ 12 = 0 3x $-$ 4y $-$ 7 = 0 $\therefore$ Equation of circle is (x2 + y2 + 4x $-$ 6y $-$ 12) + $\lambda$ (3x $-$ 4y $-$ 7) = 0 It passes through (4, 0) : (16 + 16 $-$ 12) + $\lambda$ (12 $-$ 7) = 0 $\Rightarrow$ 20 + $\lambda$ (5) = 0 $\Rightarrow$ $\lambda$ = $-$ 4 $\therefore$ (x2 + y2 + 4x $-$ 6y $-$ 12) $-$ 4(3x $-$ 4y $-$ 7) = 0 or x2 + y2 $-$ 8x + 10y + 16 = 0 Radius =

\sqrt {16 + 25 - 16} = 5

Q128

Three circles of radii a, b, c (a < b < c) touch each other externally. If they have x-axis as a common tangent, then :

A a, b, c are in A.P.

B

\sqrt a ,\sqrt b ,\sqrt c

are in A.P

C

{1 \over {\sqrt b }} + {1 \over {\sqrt c }}

=

{1 \over {\sqrt a }}

D

{1 \over {\sqrt b }} = {1 \over {\sqrt a }} + {1 \over {\sqrt c }}

Correct Answer

Option C

Solution

AB = AC + CB

\sqrt {{{\left( {b + c} \right)}^2} - {{\left( {c - b} \right)}^2}}

=

\sqrt {{{\left( {b + a} \right)}^2} - {{\left( {b - a} \right)}^2}}

+

\sqrt {{{\left( {c + a} \right)}^2} - {{\left( {c - a} \right)}^2}}

$\Rightarrow$

\sqrt {2bc}

=

\sqrt {2ac}

+

\sqrt {2ab}

Dividing by

\sqrt {abc}

we get. $\Rightarrow$

{1 \over {\sqrt a }}

=

{1 \over {\sqrt b }}

+

{1 \over {\sqrt c }}

Q129

If a circle of radius R passes through the origin O and intersects the coordinates axes at A and B, then the locus of the foot of perpendicular from O on AB is :

A (x2 + y2)2 = 4R2x2y2

B (x2 + y2) (x + y) = R2xy

C (x2 + y2)2 = 4Rx2y2

D (x2 + y2)3 = 4R2x2y2

Correct Answer

Option D

Solution

Slope of AB =

{{ - h} \over k}

Equation of AB is hx + ky = h2 + k2 A

\left( {{{{h^2} + {k^2}} \over h},0} \right),B\left( {0,{{{h^2} + {k^2}} \over k}} \right)

AB = 2R $\Rightarrow$ (h2 + k2)3 = 4R2h2k2 $\Rightarrow$ (x2 + y2)3 = 4R2x2y2

Q130

Let C1 and C2 be the centres of the circles x2 + y2 – 2x – 2y – 2 = 0 and x2 + y2 – 6x – 6y + 14 = 0 respectively. If P and Q are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral PC1QC2 is :

A 4

B 6

C 9

D 8

Correct Answer

Option A

Solution

Area = 2 $\times$

{1 \over 2}

.4 = 2