Question

The equation of the image of 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. $x^2+y^2+32x+4y+235=0$

Answer 1

Answer: (D)

The centre of the given circle is (-8, 12) and radius is 5.
The image of the circle will have the same radius, i.e. the radius of the required circle is 5.
The centre D of the required circle is the image of the centre C of the given circle in the line mirror. If D be $\left(\alpha ,\beta \right)$ then
$\frac{\alpha +8}{4}=\frac{\beta -12}{7}=-2\left[\frac{4\times -8+7\times 12+13}{4^2+7^2}\right]$ [See straight line]
Or, $\frac{\alpha +8}{4}=\frac{\beta -12}{7}=\frac{-2\times 65}{65}=-2$
$\therefore \alpha =-16,\beta =-2$
$\therefore$ Required circle is ${\left(x+16\right)}^2+{\left(y+2\right)}^2=5^2$