Question
A variable circle passes through the fixed point $$A\left( {p,\,q} \right)$$ and touches $$x$$-axis . The locus of the other end of the diameter through $$A$$ is-
A.
$${\left( {y - q} \right)^2} = 4px$$
B.
$${\left( {x - q} \right)^2} = 4py$$
C.
$${\left( {y - p} \right)^2} = 4qx$$
D.
$${\left( {x - p} \right)^2} = 4qy$$
Answer :
$${\left( {x - p} \right)^2} = 4qy$$
Solution :
Let the variable circle be
$$\eqalign{
& {x^2} + {y^2} + 2gx + 2fy + c = 0.....(1) \cr
& \therefore {p^2} + {q^2} + 2gp + 2fq + c = 0.....(2) \cr} $$
Circle (1) touches $$x$$-axis,
$$\eqalign{
& \therefore {g^2} - c = 0 \Rightarrow c = {g^2}\,.\,{\text{From }}(2) \cr
& {p^2} + {q^2} + 2gp + 2fq + {g^2} = 0.....(3) \cr} $$
Let the other end of diameter through $$\left( {p,\,q} \right)$$ be $$\left( {h,\,k} \right)$$
then, $$\frac{{h + p}}{2} = - g$$ and $$\frac{{k + q}}{2} = - f$$
Put in (3)
$$\eqalign{
& {p^2} + {q^2} + 2p\left( { - \frac{{h + p}}{2}} \right) + 2q\left( { - \frac{{k + q}}{2}} \right) + {\left( {\frac{{h + p}}{2}} \right)^2} = 0 \cr
& \Rightarrow {h^2} + {p^2} - 2hp - 4kq = 0 \cr} $$
$$\therefore $$ locus of $$\left( {h,\,k} \right)$$ is $${x^2} + {p^2} - 2xp - 4yq = 0$$
$$ \Rightarrow {\left( {x - p} \right)^2} = 4qy$$