Q. The equation of normal of $x^{2} + y^{2} - 2x + 4y - 5 = 0$ at (2, 1) is
A) $y = 3x - 5$
B) $2y = 3x - 4$
C) $y = 3x + 4$
D) $y = x + 1$
Explanation
(a)
Centre of circle is $(1,\ - 2)$
$\therefore$ Required equation of normal=equation of straight line passing through $(1,\ - 2)$ and (2, 1)
$ie,\ y + 2 = \frac{- 2 - 1}{1 - 2}(x - 1)$
$\Rightarrow \ \ y + 2 = 3x - 3$
$\Rightarrow \ \ 3x - y - 5 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.