A line is at a constant distance $c$ from the origin and meets the coordinate axes in $A$ and $B$. The locus…
A line is at a constant distance $c$ from the origin and meets the coordinate axes in $A$ and $B$. The locus of the centre of the circle passing through $O, A, B$ is
$x^2+y^2=c^2$
$x^2+y^2=2 c^2$
$x^2+y^2=3 c^2$
$x^2+y^2=4 c^2$
Solution
Let the equation of the circle be
$
x^2+y^2+2 g x+2 f y+c=0
$
It passes through origin
$
\text { so } c=0
$
Then, the equation of circle is
$
x^2+y^2+2 g x+2 f y=0
$
It also passes through $A\left(x_1, 0\right)$
$
\begin{array}{rlrl}
\therefore & x_1^2+0+2 g x_1 & =0 \\
& \Rightarrow & g & =-\frac{x_1}{2}
\end{array}
$
It also passes through $B\left(0, y_1\right)$
$
\therefore o+y_1^2+2 f y_1=0 \Rightarrow f=\frac{-y_1}{2}
$
$\Rightarrow$ centre of the circle is $\left(\frac{x_1}{2}, \frac{y_1}{2}\right)$
Mid point of $A B$ is $\left(\frac{x_1}{2}, \frac{y_1}{2}\right)$
$
\text { i.e., } O M=A M=B M
$
Thus,
$
\begin{aligned}
c=O M \Rightarrow c & =\sqrt{\frac{x_1^2}{2}+\frac{y_1^2}{2}} \\
x_1^2+y_1^2 & =4 c^2
\end{aligned}
$
Thus, locus is $x^2+y^2=4 c^2$