AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper

The equations of given circles are,
S1≡x2+y2−4=0
S2≡x2+y2−6x−8y+10=0
S3≡x2+y2+2x−4y−2=0
Let the equation of required circle be,
S≡x2+y2+2gx+2fy+c=0
The circle S=0 cuts the circles S1=0,S2=0,S3=0 atextermities of the diameter so, common chord of S=0 and S1=0 passes through center of circle S1=0 so, c=−4
Similarly (2g+6)x+(2f+8)y−14=0 passes through (3,4) so, 6g+8f+36=0
3g+4f+18=0 .....(I)
And,
(2g−2)x+(2f+4)y−2=0 passes through (-1,2) so
−2g+4f+8=0 .....(II)
Solving equation (I) and (II),
(g,f)=(−2,−3)
So, equation of circle is,
x2+y2−4x−6y−4=0