Find the equation of locus of a point P such that the distance of P from the origin is twice the distance of P from A(1, 2).

Let P(x, y) be a point on the required locus
Let origin O = (0, 0) ; Given A = (1, 2)
Given condition,
$\mathrm{OP}=2\mathrm{PA}\text{ S.O.B.S }\Rightarrow {\mathrm{OP}}^2=4{\mathrm{PA}}^2$
$\begin{array}{l}\Rightarrow (\mathrm{x}-0{)}^2+(\mathrm{y}-0{)}^2=4\left[(\mathrm{x}-1{)}^2+(\mathrm{y}-2{)}^2\right]\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2=4\left[{\mathrm{x}}^2-2\mathrm{x}+1+{\mathrm{y}}^2-4\mathrm{y}+4\right]\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2=4\left[{\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-4\mathrm{y}+5\right]\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2=4{\mathrm{x}}^2+4{\mathrm{y}}^2-8\mathrm{x}-16\mathrm{y}+20\\ \Rightarrow 4{\mathrm{x}}^2+4{\mathrm{y}}^2-8\mathrm{x}-16\mathrm{y}+20-{\mathrm{x}}^2-{\mathrm{y}}^2=0\\ \Rightarrow 3{\mathrm{x}}^2+3{\mathrm{y}}^2-8\mathrm{x}-16\mathrm{y}+20=0\end{array}$
$\therefore \text{ The locus of }\mathrm{P}\text{ is }3{\mathrm{x}}^2+3{\mathrm{y}}^2-8\mathrm{x}-16\mathrm{y}+20=0$