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1

### AIEEE 2003

MCQ (Single Correct Answer)
The lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of a circle having area as 154 sq. units. Then the equation of the circle is
A
$${x^2}\, + \,{y^2} - \,2x\, + \,2y\,\, = \,62$$
B
$${x^2}\, + \,{y^2} + \,2x\, - \,2y\,\, = \,62$$
C
$${x^2}\, + \,{y^2} + \,2x\, - \,2y\,\, = \,47$$
D
$${x^2}\, + \,{y^2} - \,2x\, + \,2y\,\, = \,47$$

## Explanation

$$\pi {r^2} = 154 \Rightarrow r = 7$$

For center on solving equation

$$2x - 3y = 5\& 3x - 4y = 7$$

we get $$x = 1,\,y = - 1$$

$$\therefore$$ center $$=(1,-1)$$

Equation of circle,

$${\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = {7^2}$$

$${x^2} + {y^2} - 2x + 2y = 47$$
2

### AIEEE 2003

MCQ (Single Correct Answer)
If the two circles $${(x - 1)^2}\, + \,{(y - 3)^2} = \,{r^2}$$ and $$\,{x^2}\, + \,{y^2} - \,8x\, + \,2y\, + \,\,8\,\, = 0$$ intersect in two distinct point, then
A
$$r > 2$$
B
$$2 < r < 8$$
C
$$r < 2$$
D
$$r = 2.$$

## Explanation

$$\left| {{r_1} - {r_2}} \right| < {C_1}{C_2}$$ for intersection

$$\Rightarrow r - 3 < 5 \Rightarrow r < 8\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

and $${r_1} + {r_2} > {C_1}{C_2},\,$$

$$r + 3 > 5 \Rightarrow r > 2\,\,\,...\left( 2 \right)$$

From $$\left( 1 \right)$$ and $$\left( 2 \right),$$ $$2 < r < 8.$$
3

### AIEEE 2002

MCQ (Single Correct Answer)
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

$$OA \le OP \le OB$$

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

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

### AIEEE 2002

MCQ (Single Correct Answer)
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 $${x^2} + {y^2} = 4{a^2}$$

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