JEE Advanced Maths: Circles (Coordinate Geometry) questions with solutions

Q1. A ray of light is incident at the point (-3, -2) on the line y = -2 (which is the tangent to the circle x² + y² = 4 at the point (0, -2)). The reflected ray from y = -2 is tangent to the circle x² + y² = 4. If the equation of the line along which the incident ray travels is Jx - Ky + 46 = 0, find the value of J + K.

1. 7
2. 17
3. 5
4. 12

Answer: 7

Tangent line is y=-2 (horizontal). The reflected ray from (-3,-2) is tangent to the circle, giving slope m=12/5 (non-trivial solution). By reflection in horizontal line, incident ray has slope -12/5. Its equation through (-3,-2) is 12x+5y+46=0, i.e., Jx-Ky+46=0 with J=12, K=-5, so J+K=7.