TS EAMCET 11-Sep-2020 Shift 2 Solved Paper

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
∴‌‌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