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        	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
one year ago

Arun
22576 Points
							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 =

one year ago
kkbisht
90 Points
							The answer is  option (b) 4x+3y+11=0Let 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 getc/$\sqrt{}$(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= -4/3we get after simple calculation the equation as 4x+3y+11=0 kkbisht

one year ago
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