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 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:
$(A)x^2+y^2+32x-4y+235=0$
$(B)x^2+y^2+32x+4y-235=0$
$(C)x^2+y^2+32x-4y-235=0$
$(D)x^2+y^2+32x+4y+235=0$



I tried to solve it reflection formula.The image of any point on the circle $(x,y)$ in the line mirror $4x+7y+13=0$ is $(x',y')$.And it is given by
$\frac{x'-x}{4}=\frac{y'-y}{7}=\frac{-2(4x+7y+13)}{16+49}$
$x'=\frac{33x-56y-104}{65}$
$y'=\frac{-56x-33y-182}{65}$
Then i got stuck.It this not the proper way to find the image of a circle in the line mirror.What should be the correct method to solve it?Please help me.Thanks.
You need to follow the following procedure:


Find the center and radius of circle.
Find the image of the center.
Put the co-ordinates of image in place of center in the eqn. $(x-a)^2+(y-b)^2=r^2$


Do it yourself and if you still have problems I will do it then only.
First of all find the value of centre of circle. Then substitute the value of x and y in the equation. Then you will get the value of centre of required circle. Then substitute the values in the general equation of circle . That's it.