The loucs of the centre of the circle which cuts orthogonally the circle `x^(2)+y^(2)-20x+4=0` and which touches x=2 is

Options

A. `y^(2)=16x+4`
B. `x^(2)=16y`
C. `x^(2)=16y+4`
D. `y^(2)=16x`

Answer

D

Explanation

Let the circle `x^(2)+y^(2)+2gx+2fy+c=0 " " ...(i)`
It cuts the circle `x^(2)+y^(2)-20x+4=0` orthogonally .
`:. 2(-10g+0xx f )=c+4`
`rArr -20g = c+4 " " ...(ii)`
Circle (i) touches the line x=2 i.e. x +0y-2=0
`:. |(-g+0-2)/(sqrt(1^(2)+0^(2)))|=sqrt(g^(2)+f^(2)-c)`
`rArr (g+2)^(2)+f^(2)-c rArr 4g +4 = f^(2)-c " "...(iii)`
Eliminating c from (ii) and (iii), we get
`-16g+4=f^(2)+4 rArr f^(2)+16g=0`
Hence, the locus of (-g, -f) is `y^(2)-16x=0`