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