Question

The equation of the image of the circle $x^2+y^2+16x-24y+183=0$ along 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. $x^2+y^2+32x+4y+235=0$

Answer 1

The correct option is D $x^2+y^2+32x+4y+235=0$

The given equation of the circle is $x^2+y^2+16x-24y+183=0$, which can be written as,

$\Rightarrow (x+8{)}^2+(y-12{)}^2=(5{)}^2$

Hence we can see the center of the circle , let's say $O(-8,12)$ and radius of the circle is $r=5$

If we mirror the image of the given circle by the line $4x+7y+13=0$, the radius of the circle won't change. But the position of center will get change. Let's assume the new center will be $O^{\prime }(\alpha ,\beta )$

By using below equation to find $O^{\prime }(\alpha ,\beta )$,

$\Rightarrow \frac{\alpha -(-8)}{4}=\frac{\beta -12}{7}=\frac{-2\left(4\right(-8)+7(12)+13)}{4^2+7^2}$

$\Rightarrow \frac{\alpha -(-8)}{4}=\frac{\beta -12}{7}=-2$

$\Rightarrow \alpha =-16,\beta =-2$

Hence new center $O^{\prime }$ is $O^{\prime }(-16,-2)$

The equation of the image of the circle through the mirror will be,

$\Rightarrow (x+16{)}^2+(y+2{)}^2=(5{)}^2$

$\Rightarrow x^2+y^2+32x+4y+235$

So correct option is $D$