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

Options

A. `x^(2)+y^(2)-4x-2y+5=0`
B. `x^(2)+y^(2)+4x+2y-5=0`
C. `x^(2)+y^(2)-4x-2y-5=0`
D. none of these

Answer

C

Explanation

The combined equation of two diameters is
`x^(2)-y^(2)-4x+2y+3=0`
`rArr (x-2)^(2)-(y-1)^(2)=0rArr (x+y-3)(x-y-1)=0`
Thus, the two diameters are x+y-3=0 and x-y-1=0 .
These 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)`
`rArr x^(2)+y^(2)-4x-2y-5=0`