The equation of the image of the circle `x^2+y^2+16x-24y+183=0` by the line mirror `4x+7y+13=0` is :

Options

A. `(x+16)^(2)+(y+2)^(2)=5^(2)`
B. `(x-16)^(2)+(y-2)^(2)=5^(2)`
C. `(x+16)^(2)+(y-2)^(2)=5^(2)`
D. `(x+16)^(2)+(y+2)^(2)=5^(2)`

Answer

A

Explanation

The equation of the given circle is
`x^(2)+y^(2)+16x-24y+183=0`
`rArr (x^(2)+16x+64)+(y^(2)-24y+144)=25`
`rArr(x+8)^(2)+(y-12)^(2)=5^(2)`
`rArr {x-(-8)}^(2)+(y-12)^(2)=5^(2) " " ...(i)`
Clearly, the centre of this circle is (-8, 12) and radius = 5.
The image of the circle (i) in the line mirror `4x+7y+13=0` is a circle whose centre is the image of the point (-8, 12) in the line mirror `4x+7y+13=0` and radius same as that of the given circle.
Let (h, k) be the image of (-8, 12) in the line mirror `4x+7y+13=0`. Then,
`(h+8)/(4)=(k-12)/(7)=(-2(4xx(-8)+7xx12+13))/((4^(2)+7^(2)))`
`rArr (h+8)/(4)=(k-12)/(7)=(-130)/(65)rArr h=-16, k=-2`
Hence, the centre of the required circle is (-16, -2) and radius equal to 5 units. So, its equation is
`(x+16)^(2)+(y+2)^(2)=5^(2)`