Q. Centre of circle whose normal’s are $x^{2} - 2xy - 3x + 6y = 0$, is
A) $\left( 3,\frac{3}{2} \right)$
B) $\left( 3,\ - \frac{3}{2} \right)$
C) $\left( \frac{3}{2},3 \right)$
D) None of these
Explanation
(a)
Given equation can be rewritten as
$x(x - 2y) - 3(x - 2y) = 0\ \Rightarrow x = 3$ And $x = 2y$ are two normals. Their intersection point is the centre $\left( 3,\frac{3}{2} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.