Find the locus of a point which moves such that the sum of the squares of its distances from $\left(0, b\right)$, $\left(0, -b\right)$ is $2r^2$.
Let $P \left(x, y\right)$ be a point on the required locus.
Distance of $P \left(x, y\right)$ from $\left(0, b\right)$ is $= \sqrt{x^2 + \left(y - b\right)^2}$
Distance of $P \left(x, y\right)$ from $\left(0, -b\right)$ is $= \sqrt{x^2 + \left(y + b\right)^2}$
As per the given problem,
$\left(\sqrt{x^2 + \left(y - b\right)^2}\right)^2 + \left(\sqrt{x^2 + \left(y + b\right)^2}\right)^2 = 2r^2$
i.e. $\;$ $x^2 + \left(y - b\right)^2 + x^2 + \left(y + b\right)^2 = 2r^2$
i.e. $\;$ $2x^2 + y^2 + b^2 - 2by + y^2 + b^2 + 2by = 2r^2$
i.e. $\;$ $2x^2 + 2y^2 + 2b^2 = 2r^2$
$\therefore \;$ The required equation of locus is
$x^2 + y^2 = r^2 - b^2$