Circle — Page 5 | JEE Mathematics

Q41

If a tangent to the circle x2 + y2 = 1 intersects the coordinate axes at distinct points P and Q, then the locus of the mid-point of PQ is :

A x2 + y2 – 4x2y2 = 0

B x2 + y2 - 2xy = 0

C x2 + y2 – 2x2y2 = 0

D x2 + y2 - 16x2y2 = 0

Correct Answer

Option A

Solution

Let the point of tangency A(cos $\theta$ , sin $\theta$ ) Equation of tangent at A, xcos $\theta$ + ysin $\theta$ = 1 $\therefore$ P(sec $\theta$ , 0) and Q(0, cosec $\theta$ ) Let M(h, k) is the mid-point of PQ. $\therefore$ h =

{{\sec\theta + 0} \over 2}

$\Rightarrow$ 2h = sec $\theta$ $\Rightarrow$ cos $\theta$ =

{1 \over {2h}}

.....(1) $\therefore$ k =

{{\cos ec\theta + 0} \over 2}

$\Rightarrow$ 2k = cosec $\theta$ $\Rightarrow$ sin $\theta$ =

{1 \over {2k}}

.....(2) From (1) and (2), sin2 $\theta$ + cos2 $\theta$ =

{\left( {{1 \over {2h}}} \right)^2} + {\left( {{1 \over {2k}}} \right)^2}

$\Rightarrow$

{1 \over {4{h^2}}} + {1 \over {4{k^2}}} = 1

$\Rightarrow$

{{{h^2} + {k^2}} \over {4{h^2}{k^2}}} = 1

$\Rightarrow$

{h^2} + {k^2} = 4{h^2}{k^2}

$\therefore$ Locus of the midpoint, x2 + y2 – 4x2y2 = 0

Q42

A circle touching the x-axis at (3, 0) and making an intercept of length 8 on the y-axis passes through the point :

A (1, 5)

B ( 2, 3)

C (3, 5)

D (3, 10)

Correct Answer

Option D

Solution

From the above figure equation of circle is (x - 3)2 + (y - 5)5 = 52 So (3, 10) will satisfy the equation.

Q43

If the angle of intersection at a point where the two circles with radii 5 cm and 12 cm intersect is 90o, then the length (in cm) of their common chord is :

A

{{13} \over 5}

B

{{60} \over {13}}

C

{{120} \over {13}}

D

{{13} \over 2}

Correct Answer

Option C

Solution

C1C2 =

\sqrt {{{12}^2} + {5^2}}

= 13 Area of

\Delta

AC1C2 =

{1 \over 2}.12.5 = {1 \over 2}.13{{AB} \over 2} \Rightarrow AB = {{120} \over {13}}units

Q44

The locus of the centres of the circles, which touch the circle, x2 + y2 = 1 externally, also touch the y-axis and lie in the first quadrant, is :

A

x = \sqrt {1 + 2y} ,y \ge 0

B

y = \sqrt {1 + 2x} ,x \ge 0

C

y = \sqrt {1 + 4x} ,x \ge 0

D

x = \sqrt {1 + 4y} ,y \ge 0

Correct Answer

Option B

Solution

Let the centre is (h, k) & radius is h (h, k > 0) OP = h + 1

\sqrt {{h^2} + {k^2}} = h + 1

\Rightarrow {h^2} + {k^2} = {h^2} + 2h + 1

\Rightarrow {k^2} = 2h + 1

$\therefore$ Locus is y2 = 2x + 1

Q45

The line x = y touches a circle at the point (1,1). If the circle also passes through the point (1, – 3), then its radius is :

A 3

B 2

C 2

\sqrt 2

D 3

\sqrt 2

Correct Answer

Option C

Solution

Equation of circle = (x – 1)2 + (y –1)2 + $\lambda$ (y – x) = 0 Which passes through (1, –3) So, 0 + 16 + $\lambda$ (–3 – 1) = 0 16 + $\lambda$ (–4) = 0 $\therefore$ $\lambda$ = 4 Now equation of circle (x – 1)2 + (y – 1)2 + 4y – 4x = 0 $\Rightarrow$ x2 + y2 – 6x + 2y + 2 = 0 radius =

\sqrt {9 + 1 - 2} = 2\sqrt 2

Q46

If the circles x2 + y2 + 5Kx + 2y + K = 0 and 2(x2 + y2) + 2Kx + 3y –1 = 0, (K

\in

R), intersect at the points P and Q, then the line 4x + 5y – K = 0 passes through P and Q, for :

A exactly two values of K

B no value of K

C exactly one value of K

D infinitely many values of K

Correct Answer

Option B

Solution

S1 $\equiv$ x2 + y2 + 5Kx + 2y + K = 0

{S_2} \equiv {x^2} + {y^2} + Kx + {3 \over 2}y - {1 \over 2} = 0

Equation of common chord is S1 – S2 = 0

\Rightarrow 4Kx + {y \over 2} + K + {1 \over 2} = 0

....(1) 4x + 5y – K = 0 …(2) (given) On comparing (1) and (2)

{{4K} \over 4} = {1 \over {10}} = {{2K + 1} \over { - 2K}}

\Rightarrow K = {1 \over {10}}

and

- 2K = 20K + 10

$\Rightarrow$ 22K = –10

\therefore K = {{ - 5} \over {11}}

So No value of K exists.

Q47

Let Z be the set of all integers,

A = \{ (x,y) \in Z \times Z:{(x - 2)^2} + {y^2} \le 4\}

B = \{ (x,y) \in Z \times Z:{x^2} + {y^2} \le 4\}

C = \{ (x,y) \in Z \times Z:{(x - 2)^2} + {(y - 2)^2} \le 4\}

If the total number of relation from A

\cap

B to A

\cap

C is 2p, then the value of p is :

A 16

B 25

C 49

D 9

Correct Answer

Option B

Solution

(x $-$ 2)2 + y2 $\le$ 4 x2 + y2 $\le$ 4 No. of points common in C1 & C2 is 5.

(0, 0), (1, 0), (2, 0), (1, 1), (1, $-$ 1) Similarly in C2 & C3 is 5.

No. of relations = 2

5 \times 5

= 225.

Q48

A rectangle is inscribed in a circle with a diameter lying along the line 3y = x + 7. If the two adjacent vertices of the rectangle are (–8, 5) and (6, 5), then the area of the rectangle (in sq. units) is :

A 72

B 84

C 56

D 98

Correct Answer

Option B

Solution

Distance beetween point A and B is = 14 unit As y coordinate of point A(-8, 5) and B(6, 5) are same.

So line AB is parallel to the x axis.

Point O is the center of the circle and OP is the perpendicular to the line AB.

And P is the mis point of line AB.

$\therefore$ Point P =

\left( {{{ - 8 + 6} \over 2},{{5 + 5} \over 2}} \right)

= (-1, 5) As OP is perpendicular to the line AB and AB is parallel to the x axis then OP is parallel to the y axis.

So x coordinate of line OP is constant. $\therefore$ x coordinate of O = -1 And point O is lies on the line 3y = x + 7 $\therefore$ 3y = -1 + 7 $\Rightarrow$ y = 2 $\therefore$ Center O = (-1, 2) $\therefore$ OP = 3 Then BC = 2OP = 6 $\therefore$ Area of rectangle = (AB)(BC) = 14 $\times$ 6 = 84

Q49

The common tangent to the circles x 2 + y2 = 4 and x2 + y2 + 6x + 8y – 24 = 0 also passes through the point :

A (6, –2)

B (4, –2)

C (–4, 6)

D (–6, 4)

Correct Answer

Option A

Solution

For this circle x 2 + y2 = 4 Center C1 = (0, 0) and radius r1 = 2 For this circle x2 + y2 + 6x + 8y – 24 = 0 Center C2 = (-3, -4) and radius r2 =

\sqrt {9 + 16 + 24}

= 7 So distance between center, C1C2 = 5 r1 + r2 = 7 |r1 - r2| = 5 As C1C2 = |r1 - r2| $\therefore$ Circles touches internally.

Here Equation of common tangent is same as common chord.

$\therefore$ S1 - S2 = 0 $\Rightarrow$ (x 2 + y2 - 4) - (x2 + y2 + 6x + 8y - 24) = 0 $\Rightarrow$ - 6x - 8y + 20 = 0 $\Rightarrow$ 3x + 4y - 10 = 0 By checking all the options you can see that, (6, –2) point satisfy the equation 3x + 4y - 10 = 0.

Q50

Let the abscissae of the two points

P

and

Q

on a circle be the roots of

x^{2}-4 x-6=0

and the ordinates of

\mathrm{P}

and

\mathrm{Q}

be the roots of

y^{2}+2 y-7=0

. If

\mathrm{PQ}

is a diameter of the circle

x^{2}+y^{2}+2 a x+2 b y+c=0

, then the value of

(a+b-c)

is _____________.

A 12

B 13

C 14

D 16

Correct Answer

Option A

Solution

Abscissae of PQ are roots of

{x^2} - 4x - 6 = 0

Ordinates of PQ are roots of

{y^2} + 2y - 7 = 0

and PQ is diameter $\Rightarrow$ Equation of circle is

{x^2} + {y^2} - 4x + 2y - 13 = 0

But, given

{x^2} + {y^2} + 2ax + 2by + c = 0

By comparison

a = - 2,b = 1,c = - 13

\Rightarrow a + b - c = - 2 + 1 + 13 = 12