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1

### AIEEE 2004

A variable circle passes through the fixed point A (p, q) and touches x-axis. The locus of the other end of the diameter through A is
A
$${(y\, - \,q)^2} = \,4\,px$$
B
$${(x\, - \,q)^2} = \,4\,py$$
C
$${(y\, - \,p)^2} = \,4\,qx$$
D
$${(x\, - \,p)^2} = \,4\,qy$$

## Explanation

Let the variable circle be

$${x^2} + {y^2} + 2gx + 2fy + c = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$$\therefore$$ $${p^2} + {q^2} + 2gp + 2fq + c = 0\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

Circle $$(1)$$ touches $$x$$-axis,

$$\therefore$$ $${g^2} - c = 0 \Rightarrow c = {g^2}.\,\,\,$$

From $$(2)$$

$${p^2} + {q^2} + 2gp + 2fq + {g^2} = 0\,\,\,\,\,\,\,\,\,...\left( 3 \right)$$

Let the other end of diameter through $$(p,q)$$ be $$(h,k),$$

then, $${{h + p} \over 2} = - g$$ and $${{k + q} \over 2} = - f$$

Put in $$(3)$$

$${p^2} + {q^2} + 2p\left( { - {{h + p} \over 2}} \right) + 2q\left( { - {{k + q} \over 2}} \right) + {\left( {{{h + p} \over 2}} \right)^2} = 0$$

$$\Rightarrow {h^2} + {p^2} - 2hp - 4kq = 0$$

$$\therefore$$ locus of $$\left( {h,k} \right)$$ is $${x^2} + {p^2} - 2xp - 4yq = 0$$

$$\Rightarrow {\left( {x - p} \right)^2} = 4qy$$
2

### AIEEE 2004

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

## Explanation

Let the variable circle is

$${x^2} + {y^2} + 2gx + 2fy + c = 0\,\,\,\,\,\,\,\,...\left( 1 \right)$$

It passes through $$(a,b)$$

$$\therefore$$ $${a^2} + {b^2} + 2ga + 2fb + c = 0\,\,\,\,\,\,\,...\left( 2 \right)$$

$$(1)$$ cuts $${x^2} + {y^2} = 4$$ orthogonally

$$\therefore$$ $$2\left( {g \times 0 + f \times 0} \right) = c - 4 \Rightarrow c = 4$$

$$\therefore$$ from $$(2)$$ $$\,\,\,{a^2} + {b^2} + 2ga + 2fb + 4 = 0$$

$$\therefore$$ Locus of center $$\left( { - g, - f} \right)$$ is

$${a^2} + {b^2} - 2ax - 2by + 4 = 0$$

or $$2ax + 2by = {a^2} + {b^2} + 4$$
3

### AIEEE 2004

If the lines 2x + 3y + 1 + 0 and 3x - y - 4 = 0 lie along diameter of a circle of circumference $$10\,\pi$$, then the equation of the circle is
A
$${x^2}\, + \,{y^2} + \,2x\, - \,2y - \,23\,\, = 0$$
B
$${x^2}\, + \,{y^2} - \,2x\, - \,2y - \,23\,\, = 0$$
C
$${x^2}\, + \,{y^2} + \,2x\, + \,2y - \,23\,\, = 0$$
D
$${x^2}\, + \,{y^2} - \,2x\, + \,2y - \,23\,\, = 0$$

## Explanation

Two diameters are along

$$2x+3y+1=0$$ and $$3x-y-4=0$$

solving we get center $$(1,-1)$$

circumference $$= 2\pi r = 10\pi$$

$$\therefore$$ $$r=5$$.

Required circle is, $${\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = {5^2}$$

$$\Rightarrow {x^2} + {y^2} - 2x + 2y - 23 = 0$$
4

### AIEEE 2003

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,

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

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

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