Question

A ray of light incident at the point (–2, –1) gets reflected from the tangent at (0, –1) to the circle x² + y² = 1. The reflected ray touches the circle. The equation of the line along which the incident ray moves, is –

• A. 4x – 3y + 11 = 0
• B. 4x + 3y + 11 = 0
• C. 3x + 4y + 11 = 0
• D. 4x + 3y + 7 = 0

Answer 1

Answer: (B)

4x + 3y + 11 = 0

Answer 2

Any line through (–2, –1) is y + 1 = m (x + 2)

It touches the circle if

$\left| \frac { 2 \mathrm {~m} - 1 } { \sqrt { 1 + \mathrm { m } ^ { 2 } } } \right|$ = 1 Ž m = 0, $\frac { 4 } { 3 }$

So PB

y + 1 = $\frac { 4 } { 3 }$ (x + 2) Ž 4x –3y + 5 = 0

A point on PB is (–5, –5) its image by

y = –1 is (–5, 3)

Hence equation of incident ray PP¢ is

y – 3 = $\frac { 3 + 1 } { - 5 + 2 }$(x + 5) Ž 4x + 3y + 11 = 0