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 $${x^2} + {p^2} - 2xp - 4yq = 0$$
$$ \Rightarrow {\left( {x - p} \right)^2} = 4qy$$