Question

A light ray gets reflected from $x=-2$. If the reflected ray touches the circle $x^{2}+y^{2}=4$ and the point of incident is $(-2,-4)$, then the equation of the incident ray is

• A. $4 y+3 x+22=0$
• B. $3 y+4 x+20=0$
• C. $4 y+2 x+20=0$
• D. $y+x+6=0$

Answer 1

Answer: (A)

Any tangent of $x^{2}+y^{2}=4$ is
$y=m x \pm 2 \sqrt{1+m^{2}}$.
If it passes through $(-2,-4)$, then
$(2 m-4)^{2}=4\left(1+m^{2}\right) $
or $ 4 m^{2}+16-16 m=4+4 m^{2}$

or $m=\infty, m=\frac{3}{4}$
Hence, the slope of the reflected ray is $3 / 4$.
Thus, the equation of the incident ray is
$(y+4)=-\frac{3}{4}(x+2) $
i.e. $ 4 y+3 x+22=0$