TS EAMCET 04 May 2019 Shift 2 Solved Paper

Let the equation of circle be,
x2+y2+2gx+2fy+c=0
Consider the figure shown below.The circle passes through (1,-1) Therefore,
2g−2f+c=−2
Now, the equation of circle whose end points of diameter are
(0,−1)‌ and ‌(−2,3)‌ is, ‌
x(x+2)+(y+1)(y−3)=0
x2+y2+2x−2y−3=0
The equations are orthogonal Hence,
2g−2f=c−3
So
−2−c‌‌=c−3
c‌‌=‌

1

2

And
g=−‌

5

2

f=−‌

5

4

Therefore, the equation of circle is,
2x2+2y2−10x−5y+1=0