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

Answer 1

Answer: (A)

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

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. $4y+3x+22=0$