The equation of the circle passing through the point (-1, 2) and having two diameters along the pair of linesx2−y2−4x+2y+3=0, is

The combined equation of two diameters isx2−y2−4x+2y+3=0⇒ (x−2)2−(y−1)2=0⇒(x+y−3)(x−y−1)=0Thus, the two diameters are x + y - 3 = 0 and x - y - l = CThese two intersect at (2, 1).Thus, the required circle has its centre at (2, 1) and passes through (-1, 2). So, its equation is(x−2)2+(y−1)2=(2+1)2+(1−2)2⇒ x2+y2−4x−2y−5=0