Suggested languages for you:

Americas

Europe

Problem 13

# A ray of light incident at the point $$(3,1)$$ gets reflected from the tangent at $$(0,1)$$ to the cirde $$x^{2}+y^{2}=1$$. The reflected ray touches the circle. The equation of the line along which the incident ray moves is: (a) $$3 x+4 y-13=0$$ (b) $$4 x-3 y-13=0$$ (c) $$3 x-4 y+13=0$$ (d) $$4 x-3 y-10=0$$

Expert verified
The equation of the line along which the incident ray moves is $$3x+4y-13=0$$.
See the step by step solution

## Step 1: Find the slope of tangent

The first step is to find the slope of the tangent line. The line connecting the center of circle, (0,0), and the point of tangency, (0,1), is vertical. So, the slope of this radial line is undefined. As the tangent line is perpendicular to this radial line, its slope will be 0. In mathematical terms, the slope of tangent, $$m_{t}$$, is 0.

## Step 2: Equation of the tangent line

The equation of the tangent line can be written using the point-slope form $$y-y_{1}=m_{t}(x-x_{1})$$. Substituting the point (0,1) and slope $$m_{t}=0$$ into this equation, the equation of the tangent line is $$y=1$$.

## Step 3: Find the equation of normal line

The normal line at the point of incidence (0,1), is perpendicular to the tangent line. Since the tangent line has slope 0, the normal has a slope which is undefined. Hence the equation of normal line is $$x=0$$.

## Step 4: Find the angle of incidence

As the normal line is vertical, the angle of incidence, $$i$$, is the angle the line from the point of incidence and the given point, (3,1) makes with the normal line. This angle can also be found using the slope of the line. The slope, $$m_{i}$$, is $$(y_{2}-y_{1})/(x_{2}-x_{1}) = (1-1)/(0-3) = 0$$. Hence, the angle of incidence, $$i = \arctan(m_{i}) = 0$$.

## Step 5: Apply Angle of incidence = Angle of reflection

The angle of reflection is also equal to the angle of incidence, $$r = i = 0$$. Therefore, the line of reflected ray is parallel to the normal line, and hence it is vertical passed through (3,1). This gives the equation $$x=3$$. The line of incident ray must be such that when extended it must touch the circle while maintaining the property that the angle of incidence equals the angle of reflection. Consider the various options given in the problem statement one by one.

## Step 6: Test the given options

Plug in the reflected point (3,1) in the options provided in the problem. The one which satisfies is the one we are looking for. By testing each option, we find that $$3x+4y-13=0$$ is valid. Hence, this is the equation of the line along which the incident ray moves.

We value your feedback to improve our textbook solutions.

## Access millions of textbook solutions in one place

• Access over 3 million high quality textbook solutions
• Access our popular flashcard, quiz, mock-exam and notes features

## Join over 22 million students in learning with our Vaia App

The first learning app that truly has everything you need to ace your exams in one place.

• Flashcards & Quizzes
• AI Study Assistant
• Smart Note-Taking
• Mock-Exams
• Study Planner