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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
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Verified Answer
The correct answer is:
$(x-p)^2=4 q y$
$(x-p)^2=4 q y$
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
$$
\begin{aligned}
& \Rightarrow x^2-(h+p) x+(h p+k q)=0 \Rightarrow(h+p)^2=4(h p+k q) \quad(D=0) \\
& \Rightarrow(x-p)^2=4 q y .
\end{aligned}
$$
$$
(x-h)(x-p)+(y-k)(y-q)=0
$$
Put $y=0$, since $x$-axis touches the circle
$$
\begin{aligned}
& \Rightarrow x^2-(h+p) x+(h p+k q)=0 \Rightarrow(h+p)^2=4(h p+k q) \quad(D=0) \\
& \Rightarrow(x-p)^2=4 q y .
\end{aligned}
$$
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