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 ??                                             Options  a) 4x -3y +11=0  b) 4x+3y+11=0  c) 3x +4y+11=0

Arun
25757 Points
4 years ago
Let the equation of incident ray be
y=m1x c1,
and, reflected ray be,
y=m2x c2,

Equation of tangent to circle at(0,-1),
is y=-1, slope 0,
now this tangent passes thru (-2,-1),
and the ray is incidnt at (-2,-1),
Normal at this pt to the tangent is x=-2, slope infinity,
now we know, angle of incidnc = angle of reflection,
So,
Lines y=m1x C1
and y=m2x c2
are equally inclind to normal at (-2,-1) x=-2!!

Now using formula for slope,equally inclind lines.

(m1-m)/1 m.m1 = (m-m2)/1 m.m2

Where, m=slope of normal x=-2, which is infinity,

Solving this, we get, m1 m2 =0!

Now since the reflected ray,
y=m2x c2,
is a tangent to circle,
using tangency conditon,
(c2)^2 = 1 (m2)^2,.....(1)
and also
since it passes thru (-2,-1),
c2=m2-2!.......(1)

Solving(1) and (2)
We get, m2=3/4,

But we know m1 m2=0,(provd above)

So, m1= -3/4;

Now since
y=m1x c1 also passes thru (-2,-1),
Put in equation, we get
c1=m1-2
So, c1= -11/4,

Now putting c1 and m1 in equation of incident ray,
we get, the equation as,
4y 3x 11 =
kkbisht
90 Points
4 years ago
The answer is  option (b) 4x+3y+11=0
Let y=mx+c be the reflected ray  touching (tangent) the circle .Using the condition that this reflected ray is tangent if the pependicular drawn from the centre(0,0) of the unit Circle x2 + y2=1 we get
c/$\sqrt{}$(1+m2) = 1 => c=$\sqrt{}$(1+m2) therefore the equatio is y=mx+$\sqrt{}$(1+m2). As it passes thogt the point of incidence(-2,-1)

we have     -1=m(-2) +$\sqrt{}$(1+m2). =>(2m-1)2=1+m2 =>4m2 +1-4m=1+m2 => 3m2-4m=0 => m=0 or m=4/3.
Now just see If this tangent makes an angle $\alpha }$ with the normal at the point of incidence.then this tangent will make an angle 90-$\alpha }$ with tanget at the point(0,-1) which is parrale to x-axis.
Then by definition slope  tan(90-$\alpha }$)=4/3 or cot $\alpha }$= 4/3.
Now the incident ray makes an angle 90+$\alpha }$ with the x-axis therefore its slope is tan(90+$\alpha }$)=  -cot$\alpha }$= -4/3 ( from above)
Hence the equation of the incident ray is y=mx+c passing through (-2,-1) and slope m= -4/3
we get after simple calculation the equation as 4x+3y+11=0

kkbisht