Let point P is (h,k). Given Q(0,2) and R(−1,0) then given PQ=
1
√2
PR Squaring on both sides, 2(PQ)2=(PR)2 2(h2+(k−2)2)=(h+1)2+k2
2h2+2k2−8k+8=h2+1+2h+k2
∴h2+k2−2h−8k+7=0 ∴ Locus of point P x2+y2−2x−8y+7=0....(i) ∴ Eq. (i) is circle. ∴ Centre (1,4) and radius =√1+16−7=√10 Hence, locus at point P is circle with centre (1,4) and radius √10