Equation of the circle which cuts the circle `x^2+y^2+2x+ 4y -4=0` and the lines `xy -2x -y+2=0` orthogonally, is

Options

A. `x^(2)+y^(2)-2x-4y-6=0`
B. `x^(2)+y^(2)-2x-4y+6=0`
C. `x^(2)+y^(2)-2x-4y-12=0`
D. none of these

Answer

1

Explanation

`xy-2x-y+2=0`
`implies (x-1)(y-2)=0`
`implies x=1 ` and `y=2`
Let the equation of the required circle be
`x^(2)+y^(2)+2gx+2fy+c=0`
So, centre is (1,2) (as normal intersects at centre of the circle ).
i.e., `-f =1 ` and `-f = 2`
Thus, equation of circle is `x^(2)+y^(2)-2x-4y+c=0`
This circle intersects the given circle `x^(2)+y^(2)+2x+4y-4=0` orthogonally
`:. 2(-1)(1)+2(-2)(2)= c -4`
`implies c= -6`
Hence, required circle is `x^(2)+y^(2)-2x-4y-6=0`