AP EAMCET Engineering 23 Apr 2019 Shift 1 Paper

The given equation of circle is,
(x+a)2+(y+b)2=a2
x2+y2+2ax+2by+b2=0 ....(I)
And,
(x+c)2+(y+d)2=a2
x2+y2+2cx+2dy+c2=0 ....(II)
From equation (I),
g1=a,f1=b,c1=b2
From equation (II),
g2=c,f2=d,c2=c2
If circle are orthogonal then,
2[g1 g2+f1f2 ]=c1+c2
2(ac+bd)=b2+c2
2ac+2bd=b2+c2
2ac−c2=b2−2bd
Further simplify the above,
c(2a−c)=b(b−2d)
b(b−2d)=c(2a−c)