Question

A variable circle passes through the point $P(1,2)$ and touches the $x-$axis. The locus of the other end of the diameter through $P$ is

• A. $(x-1{)}^2=8y$
• B. $(x+1{)}^2=8y$
• C. $(y-1{)}^2=8x$
• D. $(x-1{)}^2+8y=0$

Answer 1

The correct option is A $(x-1{)}^2=8y$

Equation of any circle touching $x-$axis is of the form
$(x-h{)}^2+(y-k{)}^2=k^2\cdots \left(1\right)$
Let the coordinates of the other end of the diameter through $P$ be $(\alpha ,\beta )$
Then, $\frac{\alpha +1}{2}=h$ and $\frac{\beta +2}{2}=k$

Also, putting $P(1,2)$ in $\left(1\right),$ we get
$(1-h{)}^2+(2-k{)}^2=k^2$
or $(h-1{)}^2+(k-2{)}^2=k^2$
$\Rightarrow {(\frac{\alpha +1}{2}-1)}^2+{(\frac{\beta +2}{2}-2)}^2={\left(\frac{\beta +2}{2}\right)}^2$
$\Rightarrow (\alpha -1{)}^2+(\beta -2{)}^2=(\beta +2{)}^2$
$\Rightarrow (\alpha -1{)}^2=8\beta$
$\therefore$ Locus of $(\alpha ,\beta )$ is $(x-1{)}^2=8y$