Find the equation of the image of the circle `x^2+y^2+8x-16y+64=0` in the line mirror `x=0.`

Options

A. `x^(2)+y^(2)-8x-16y+64=0`
B. `x^(2)+y^(2)-8x+16y+64=0`
C. `x^(2)+y^(2)+8x+16y+64=0`
D. `x^(2)+y^(2)+8x-16y+64=0`

Answer

A

Explanation

The equation of the given circle is
`x^(2)+y^(2)+8x-16y+64=0`
`rArr (x^(2)+8x+16)+(y^(2)-16y+64)=16`
`rArr (x+4)^(2)+(y-8)^(2)=4^(2)`
`rArr {x-(-4)}^(2)+(y-8)^(2)=4^(2)`
Clearly, its centre is at (-4, 8) and radius = 4.
The image of this circle in the line mirror has its centre `C_(1) (4, 8)` and radius 4. So, its equation is
`(x+4)^(2)+(y-8)^(2)=4^(2)` or, `x^(2)+y^(2)-8x-16y+64=0`