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1

### AIEEE 2005

If the pair of lines $$a{x^2} + 2\left( {a + b} \right)xy + b{y^2} = 0$$ lie along diameters of a circle and divide the circle into four sectors such that the area of one of the sectors is thrice the area of another sector then
A
$$3{a^2} - 10ab + 3{b^2} = 0$$
B
$$3{a^2} - 2ab + 3{b^2} = 0$$
C
$$3{a^2} + 10ab + 3{b^2} = 0$$
D
$$3{a^2} + 2ab + 3{b^2} = 0$$

## Explanation

As per question area of one sector $$=3$$ area of another sector

$$\Rightarrow$$ at center by one sector $$= 3 \times$$ angle at center by another sector

Let one angle be $$\theta$$ then other $$=30$$

Clearly $$\theta + 3\theta = 180 \Rightarrow \theta = {45^ \circ }$$

$$\therefore$$ Angle between the diameters represented by combined equation

$$a{x^2} + 2\left( {a + b\,\,\,xy} \right) + b{y^2} = 0$$ is $${45^ \circ }$$

$$\therefore$$ Using $$tan$$ $$\theta$$ $$= {{2\sqrt {{h^2} - ab} } \over {a + b}}$$

we get $$\tan \,{45^ \circ } = {{2\sqrt {{{\left( {a + b} \right)}^2} - ab} } \over {a + b}}$$

$$\Rightarrow 1 = {{2\sqrt {{a^2} + {b^2} + ab} } \over {a + b}}$$

$$\Rightarrow {\left( {a + b} \right)^2} = 4\left( {{a^2} + {b^2} + ab} \right)$$

$$\Rightarrow {a^2} + {b^2} + 2ab = 4{a^2} + 4{b^2} + 4ab$$

$$\Rightarrow 3{a^2} + 3{b^2} + 2ab = 0$$
2

### AIEEE 2005

If the circles $${x^2}\, + \,{y^2} + \,2ax\, + \,cy\, + a\,\, = 0$$ and $${x^2}\, + \,{y^2} - \,3ax\, + \,dy\, - 1\,\, = 0$$ intersect in two ditinct points P and Q then the line 5x + by - a = 0 passes through P and Q for
A
exactly one value of a
B
no value of a
C
infinitely many values of a
D
exactly two values of a

## Explanation

$${s_1} = {x^2} + {y^2} + 2ax + cy + a = 0$$

$${s_2} = {x^2} + {y^2} - 3ax + dy - 1 = 0$$

Equation of common chord of circles $${s_1}$$ and $${s_2}$$ is

given by $${s_1} - {s_2} = 0$$

$$\Rightarrow 5ax + \left( {c - d} \right)y + a + 1 = 0$$

Given that $$5x + by - a = 0$$ passes through $$P$$ and $$Q$$

$$\therefore$$ The two equations should represent the same line

$$\Rightarrow {a \over 1} = {{c - d} \over b} = {{a + 1} \over { - a}}$$

$$\Rightarrow a + 1 = - {a^2}$$

$${a^2} + a + 1 = 0$$

No real value of $$a.$$
3

### AIEEE 2005

If a circle passes through the point (a, b) and cuts the circle $${x^2}\, + \,{y^2} = {p^2}$$ orthogonally, then the equation of the locus of its centre is
A
$${x^2}\, + \,{y^2} - \,3ax\, - \,4\,by\,\, + \,({a^2}\, + \,{b^2} - {p^2}) = 0$$
B
$$2ax\, + \,\,2\,by\,\, - \,({a^2}\, - \,{b^2} + {p^2}) = 0$$
C
$${x^2}\, + \,{y^2} - \,2ax\, - \,\,3\,by\,\, + \,({a^2}\, - \,{b^2} - {p^2}) = 0$$
D
$$2ax\, + \,\,2\,by\,\, - \,({a^2}\, + \,{b^2} + {p^2}) = 0$$

## Explanation

Let the center be $$\left( {\alpha ,\beta } \right)$$

As It cuts the circle $${x^2} + {y^2} = {p^2}$$ orthogonally

$$\therefore$$ Using $$2{g_1}{g_2} + 2{f_1}{f_2} = {c_1} + {c_2},\,\,$$ we get

$$2\left( { - \alpha } \right) \times 0 + 2\left( { - \beta } \right) \times 0$$

$$= {c_1} - {p^2} \Rightarrow {c_1} = {p^2}$$

Let equation of circle is

$${x^2} + {y^2} - 2\alpha x - 2\beta y + {p^2} = 0$$

It passes through

$$\left( {a,b} \right) \Rightarrow {a^2} + {b^2} - 2\alpha a - 2\beta b + {p^2} = 0$$

$$\therefore$$ Locus of $$\left( {\alpha ,\beta } \right)$$ is

$$\therefore$$ $$2ax + 2by - \left( {{a^2} + {b^2} + {p^2}} \right) = 0.$$
4

### AIEEE 2005

A circle touches the x-axis and also touches the circle with centre at (0, 3) and radius 2. The locus of the centre of the circle is
A
an ellipse
B
a circle
C
a hyperbola
D
a parabola

## Explanation

Equation of circle with center $$(0,3)$$ and radius $$2$$ is

$${x^2} + {\left( {y - 3} \right)^2} = 4$$

Let locus of the variable circle is $$\left( {\alpha ,\beta } \right)$$

As it touches $$x$$-axis.

$$\therefore$$ It's equation is $${\left( {x - \alpha } \right)^2} + {\left( {y + \beta } \right)^2} = {\beta ^2}$$

Circle touch externally $$\Rightarrow {c_1}{c_2} = {r_1} + {r_2}$$

$$\therefore$$ $$\sqrt {{\alpha ^2} + {{\left( {\beta - 3} \right)}^2}} = 2 + \beta$$

$${\alpha ^2} + {\left( {\beta - 3} \right)^2} = {\beta ^2} + 4 + 4\beta$$

$$\Rightarrow {\alpha ^2} = 10\left( {\beta - 1/2} \right)$$

$$\therefore$$ Locus is $${x^2} = 10\left( {y - {1 \over 2}} \right)$$ which is parabola.

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