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[g1g2+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)