Question

Find the equation of the normal to the circle x² + y² – 5x + 2y – 48 = 0 at the point (5, 6).

Answer 1

The equation of the tangent to the circle x² + y² – 5x + 2y – 48 = 0 at (5, 6) is

 ⇒ 10x + 12y – 5x – 25 + 2y + 12 – 96 = 0 

⇒ 5x + 14y – 109 = 0

 ∴ Slope of the tangent = - 5/14

⇒ Slope of the norma = 14/5

Hence, the equation of the normal at (5, 6) is 

y – 6 = (14/5) (x – 5) 

⇒ 14x – 5y – 40 = 0