NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

### AIEEE 2002

The centres of a set of circles, each of radius 3, lie on the circle $${x^2}\, + \,{y^2} = 25$$. The locus of any point in the set is
A
$$4\, \le \,\,{x^2}\, + \,{y^2}\, \le \,\,64$$
B
$${x^2}\, + \,{y^2}\, \le \,\,25$$
C
$${x^2}\, + \,{y^2}\, \ge \,\,25$$
D
$$3\, \le \,\,{x^2}\, + \,{y^2}\, \le \,\,9$$

## Explanation

For any point $$P(x,y)$$ in the given circle,

we should have

$$OA \le OP \le OB$$

$$\Rightarrow \left( {5 - 3} \right) \le \sqrt {{x^2} + {y^2}} \le 5 + 3$$

$$\Rightarrow 4 \le {x^2} + {y^2} \le 64$$
2

### AIEEE 2002

The equation of a circle with origin as a center and passing thorough 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
$${x^2}\, + \,{y^2} = {a^2}$$

## Explanation

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 $${x^2} + {y^2} = 4{a^2}$$
3

### AIEEE 2002

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

## Explanation

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

### AIEEE 2002

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

## Explanation

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 $${45^ \circ }.$$

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

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

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

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

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

$$= - 1 \pm \sqrt 2 .$$

### Joint Entrance Examination

JEE Main JEE Advanced WB JEE

### Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

NEET

Class 12