The equation of the circle which touches the circle `x^2+y^2-6x+6y+17 = 0` externally and to which the lines `x^2-3 xy-3x + 9y = 0` are normals, is

Options

A. `x^(2)+y^(2)-6x-2y-1=0`
B. `x^(2)+y^(2)-6x-2y+1=0`
C. `x^(2)+y^(2)+6x+2y+1=0`
D. `x^(2)+y^(2)-6x+2y+1=0`

Answer

B

Explanation

The combined equation of the normals to the given circle is
`x^(2)-3xy-3x+9y=0`
`rArr x(x-3y)-3(x-3y)=0`
`rArr (x-3y)(x-3)=0rArr x-3y=0, x=3`.
So, the equations of the normals to the required circle are x-3y=0 and x=3. These two normals intersect at (3, 1).
Therefore, coordinates of the centre of the required circle are (3, 1). Its radius is 3.
Thus, the equation of the required circle is
`(x-3)^(2)+(y-1)^(2)=3^(2)` or, `x^(2)+y^(2)-6x-2y+1=0`