The equation of the circle passing through the point of intersection of the circles `x^2+y^2-4x-2y=8` and `x^2+y^2-2x-4y=8` and the point `(-1,4)` is (a) `x^2+y^2+4x+4y-8=0` (b) `x^2+y^2-3x+4y+8=0` (c) `x^2+y^2+x+y=0` (d) `x^2+y^2-3x-3y-8=0`

Options

A. `x^(2)+y^(2)+4x+4y-8=0`
B. `x^(2)+y^(2)-3x+4y+8=0`
C. `x^(2)+y^(2)+x+y-8=0`
D. `x^(2)+y^(2)-3x-3y-8=0`

Answer

4

Explanation

The equation of any circle through the points of intersection of the given circles is
`x^(2)+y^(2)-4x-2y-8+k(x^(2)+y^(2)-2x-4y-8)=0` (1)
Since circle (1) passes through `(-1,4)`, we have
`k=1`
Therefore, the requiredcircle is
`x^(2)+y^(2)-3x-3y-8=0`