TS EAMCET 4 May 2018 Shift 1 Solved Paper

Let the equation of circle be,
x2+y2+2gx+2fy+c=0
The circle passes through the point (1,0)
Hence,
1+2g+c=0
The circle is cutting the circles x2+y2−2x+4y+1=0 and, x2+y2+6x−2y+1=0 orthogonally.
So,
2g(−1)+2f(2)=c+1
And,
2g(3)+2f(−1)=c+1
Therefore,
f=0
g=0
And,
c=−1
Hence, the equation of required circle is x2+y2=1 and its centre is (0,0)