Question

The equation of the image of the circle $x^2+y^2+16x-24y+183=0$ by the 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

$x^2+y^2+32x+4y+235=0$

$x^2+y^2+16x-24y+183=0$

Centre $\equiv$ (-8, 12) Radius = 5

Let $(x_1,y_1)$ be the image of (-8, 12) w.r.t. to the line

$4x+7y+13=0$

$\therefore \frac{x_1-(-8)}{4}=\frac{y_1-12}{7}$

$=\frac{-2\{4\times (-8)+7\times 12+13\}}{4^2+7^2}$

$\frac{x_1+8}{4}=\frac{y_1-12}{7}=-2$

$x_1=-16,y_1=-2$

Equation of required circle is

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

$x^2+y^2+32x+4y+235=0$