Find the equation of the locus of a point whose distance from $\left(2, -3\right)$ is double its distance from $\left(1, 2\right)$.
Let the given points be $\;$ $A \left(2, -3\right)$ $\;$ and $\;$ $B \left(1, 2\right)$.
Let $P \left(x, y\right)$ be a point on the given locus.
Distance $\;$ $PA = \sqrt{\left(x - 2\right)^2 + \left(y + 3\right)^2}$
Distance $\;$ $PB = \sqrt{\left(x - 1\right)^2 + \left(y - 2\right)^2}$
As per question, $\;$ $PA = 2 \times PB$
i.e. $\;$ $\left(PA\right)^2 = 4 \times \left(PB\right)^2$
i.e. $\;$ $\left(x - 2\right)^2 + \left(y + 3\right)^2 = 4 \times \left[\left(x - 1\right)^2 + \left(y - 2\right)^2\right]$
i.e. $\;$ $x^2 - 4x + 4 + y^2 + 6y + 9 = 4 \times \left[x^2 - 2x + 1 + y^2 - 4y + 4\right]$
i.e. $\;$ $x^2 + y^2 -4x + 6y + 13 = 4x^2 + 4y^2 -8x - 16y + 20$
i.e. $\;$ $3x^2 + 3y^2 -4x -22y +7 = 0$ $\;\;\;$ is the equation of the required locus.