Let the equation of one of the circles be x2+y2+2gx+2fy+c=0 Since, it passes through (a,b). Hence, a2+b2+2ga+2fb+c=0 Also, it cuts the circle x2+y2−2x+4y−4=0 orthogonally. Then −2g(1)−2f(−2)=c+(−4) ⇒−2g+4f=c−4 ⇒−2g+4f=−a2−b2−2ga−2fb−4 ⇒a2+b2+(−2+2a)g Thus, the locus of the centre (−g,−f) is (a2+b2+4)+2(−1+a)x+2(2+b)y=0 ⇒(a−1)x+(b+2)y=−(