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1

### 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.
2

### AIEEE 2004

Intercept on the line y = x by the circle $${x^2}\, + \,{y^2} - 2x = 0$$ is AB. Equation of the circle on AB as a diameter is
A
$$\,{x^2}\, + \,{y^2} + \,x\, - \,y\,\, = 0$$
B
$$\,{x^2}\, + \,{y^2} - \,x\, + \,y\,\, = 0$$
C
$$\,{x^2}\, + \,{y^2} + \,x\, + \,y\,\, = 0$$
D
$$\,{x^2}\, + \,{y^2} - \,x\, - \,y\,\, = 0$$

## Explanation

Solving $$y=x$$ and the circle

$${x^2} + {y^2} - 2x = 0,$$ we get

$$x = 0,y = 0$$ and $$x=1,$$ $$y=1$$

$$\therefore$$ Extremities of diameter of the required circle are

$$\left( {0,0} \right)$$ and $$\left( {1,1} \right)$$. Hence, the equation of circle is

$$\left( {x - 0} \right)\left( {x - 1} \right) + \left( {y - 0} \right)\left( {y - 1} \right) = 0$$

$$\Rightarrow {x^2} + {y^2} - x - y = 0$$
3

### 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 $${x^2} + {p^2} - 2xp - 4yq = 0$$

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

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

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