Question

The equation of the image of the circle $x^2+y^2+32x+4y+235=0$ by the line mirror $4x+7y+l3=0$ is

• A. $x^2+y^2+32x-4y+235=0$
• B. $x^2+y^2+32x-4y-235=0$
• C. $x^2+y^2+16x-24y+183=0$
• D. $x^2+y^2-16x+24y+183=0$

Answer 1

The correct option is C $x^2+y^2+16x-24y+183=0$

$P(x_1,y_1)$ be the centre of the required image circle.

$\therefore$ $P(x_1,y_1)$ is the image of the centre $(-16,-2)$ of the given circle.

$\therefore$ $\left(\frac{x_1-16}{2},\frac{y_1-2}{2}\right)$ lies on the line $4x+7y+13=0$

$\Rightarrow \frac{4(x_1-16)}{2}+\frac{7(y_1-2)}{2}+13=0$

i. e., $4x_1+7y_1=52$ ... (1)

Also the line joining the two centres is perpendicular to $4x+7y+13=0$.

$\therefore$ $(-\frac{4}{7}\left)\right(\frac{y_1+2}{x_1+16})=-1$

i.e., $7x_1-4y_1=-104$ ... (2)

Solving (1) and (2)

$65x_1=-520\Rightarrow x_1=-8$ and $y_1=12$

Radius of the image circle $=$ radius of the original circle
$=\sqrt{{16}^2+2^2-235}=5$

$\therefore$ the equation of the image circle is $(x+8{)}^2+(y-12{)}^2=25$
$i.e.,x^2+y^2+16x-24y+183=0$.