Q125 / 225 · JEE Main 2004 · Mathematics
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 $\mathrm{A}$ is
- Correct answerA$(x-p)^2=4 q y$
- B$(x-q)^2=4 p y$
- C$(y-p)^2=4 q x$
- D$(y-q)^2=4 p x$
Solution
Let the other end of diameter is $(\mathrm{h}, \mathrm{k})$ then equation of circle is
$$
(x-h)(x-p)+(y-k)(y-q)=0
$$
Put $y=0$, since $x$-axis touches the circle
$$
\begin{aligned}
& \Rightarrow x^2-(h+p) x+(h p+k q)=0 \Rightarrow(h+p)^2=4(h p+k q) \quad(D=0) \\
& \Rightarrow(x-p)^2=4 q y .
\end{aligned}
$$
Hence, the correct answer is option: 'A'.