Find the equation of the locus of a point such that the sum of the squares of whose distances from the points $\left(2, 4\right)$ and $\left(-3, -1\right)$ is $30$.
Let $\;$ $A = \left(2, 4\right)$, $\;$ $B = \left(-3, -1\right)$
Let $P \left(x, y\right)$ be a point on the required locus.
Distance $\;$ $PA = \sqrt{\left(x - 2\right)^2 + \left(y - 4\right)^2}$
i.e. $\;$ $PA = \sqrt{x^2 + y^2 - 4x - 8y + 20}$
$\therefore \;$ $\left(PA\right)^2 = x^2 + y^2 - 4x - 8y + 20$
Distance $\;$ $PB = \sqrt{\left(x + 3\right)^2 + \left(y + 1\right)^2}$
i.e. $\;$ $PB = \sqrt{x^2 + y^2 + 6x + 2y + 10}$
$\therefore \;$ $\left(PB\right)^2 = x^2 + y^2 + 6x + 2y + 10$
As per question, $\;$ $\left(PA\right)^2 + \left(PB\right)^2 = 30$
i.e. $\;$ $x^2 + y^2 - 4x - 8y + 20 + x^2 + y^2 + 6x + 2y + 10 = 30$
i.e. $\;$ $2 x^2 + 2 y^2 + 2x - 6y = 0$
i.e. $\;$ $x^2 + y^2 + x - 3y = 0$ $\;\;\;$ is the equation of the required locus.