TS EAMCET 5 May 2018 Shift 2 Solved Paper

It is given that P is the mid-point of QR and coordinate of point P be (h,k).
Therefore,
h=

x1+1

2

x1=2h−1
And,
k=

y1

2

y1=2k
Now (x1,y1) lies on the circle x2+y2=1
Therefore,
(2h−1)2+(2k)2=1
h2+k2−h=0
Thus, the locus of the point P is,
x2+y2−x=0