A point  $P$  moves so that distance from $(0,2)$ to $P$  is $\frac{1}{\sqrt{2}}$  times the distance of  $P$ from $(-1,0)$. Then the locus of the point is

Let point P is  $(h,k)$. Given $Q(0,2)$  and  $R(-1,0)$  then given   $PQ=\frac{1}{\sqrt{2}}PR$
Squaring on both sides,
  $$\begin{gathered} 2{\left(PQ\right)}^2={\left(PR\right)}^2 \\ 2(h^2+{(k-2)}^2)={(h+1)}^2+k^2 \\ 2h^2+2k^2-8k+8=h^2+1+2h+k^2 \\ \therefore h^2+k^2-2h-8k+7=0 \end{gathered}$$
 $\therefore$  Locus of point P
  $x^2+y^2-2x-8y+7=0\mathrm{.......}\left(i\right)$
  $\therefore$  Equation (i) is circle.
 $\therefore$  Centre $(1,4)$  and radius  $=\sqrt{1+16-7}=\sqrt{10}$
Hence, locus at point P is circle with centre  $(1,4)$ and radius  $\sqrt{10}$