Question

A ray of light, incident at the point $(-2,-1)$, gets reflected from the tangent at $(0,-1)$ to the circle $x^2+y^2=1$. The reflected ray touches the circle. The equation of the line along which the incident ray moved is

• A. $4x-3y+11=0$
• B. $4x+3y+11=0$
• C. $3x+4y+11=0$
• D. None of these

Answer 1

The correct option is B $4x+3y+11=0$

The equation of the reflected ray is

$(y+1)=m(x+2)$ or, $mx-y+2m-1=0$ ...(1)

Since it touches the circle $x^2+y^2=1.$

$\therefore$ length of $\perp$ from $(0,0)$ on $\left(1\right)=$ radius $=1$

$\Rightarrow \frac{|m\left(0\right)-0+2m-1|}{\sqrt{1+m^2}}=1\Rightarrow \frac{2m-1}{\sqrt{1+m^2}}=\pm 1$

$\Rightarrow {(2m-1)}^2={(1+m)}^2\Rightarrow 3m^2-4m=0\Rightarrow m=0,\frac{4}{3}.$

$\therefore$ Equation of the reflected ray is

$(y+1)=\frac{4}{3}(x+2)$ or $4x-3y+5=0.$

Let $\alpha$ be the angle between the reflected the reflected ray and the line $y=-1$

Then, $\tan \alpha =\frac{|\frac{4}{3}-0|}{1+\frac{4}{3}.0}=\pm \frac{4}{3}$

$\therefore$ Slope of the incident ray $=-\frac{4}{3}.$

Hence, equation of the incident ray is $(y+1)=\frac{-4}{3}(x+2)$

i.e., $3(y+1)=-4(x+2)\Rightarrow 4x+3y+11=0.$