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^(2)+y^(2)+32x-4y+235=0`
B. `x^(2)+y^(2)+32x+4y-235=0`
C. `x^(2)+y^(2)+32x-4y-235=0`
D. `x^(2)+y^(2)+32x+4y+235=0`

Answer

D

Explanation

Find image of centre of given circle
Given circle is
`x^(2)+y^(2)+16x-24y+183=0` ..(1)
Centre `=(-8,12):r=sqrt(64+144-183)=sqrt(25)=5`
given line is `4x+7y+13=0` .(2)
Let `(alpha,beta)` be the image of `(-8,12)` wrt (2).
`(alpha+8)/(4)=(beta-12)/(7)=-(2(4(-8)+7(12)+13))/(4^(2)+7^(2))`
`implies(alpha+beta)/(4)=(beta-12)/(7)=-(2(65))/(65)=-2`
`impliesalpha=-8-8, beta=-14+12`
`impliesalpha=-16, beta=-2`
`because` Equation of required circles is `(x+16)^(2)+(y+2)^(2)=5^(2)`
`impliesx^(2)+y^(2)+32x+4y+235=0`