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. $3x+4y+11=0$
• B. $3x+4y-11=0$
• C. $4x+3y+11=0$
• D. $4x-3y+11=0$

Answer 1

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

So, $e{q}^n$ of tangent at $(0,1)$ is
$y=-1$
So, $e{q}^n$ of tangent from $(0,-1)$ to circle $x^2+y^2=1$ is
$y+1=m(x+2)$
$y=mx+2m-1$
So, $1=\left|\frac{2m-1}{\sqrt{1+m^2}}\right|$
$1+m^2-4m=0$
$3m^2-4m=0$
$m=0$ or $m=\frac{4}{3}$
So, stope of $PB=\frac{4}{3}$
then $tan\theta =\left|\frac{m-0}{1+m\left(0\right)}\right|=\left|\frac{0-\frac{4}{3}}{1+101\times \frac{4}{3}}\right|$
$m=|-\frac{4}{3}|$
$\therefore m=-\frac{4}{3}$
So, $e{q}^n$ of incident ray is
$y+1=\frac{-4}{3}(x+2)$
$3y+3=-4x-8$
$\Rightarrow 3y+4x+11=0$