Q. Centre of circle whose normals 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
$x(x - 2y) - 3(x - 2y) = 0$
$\Rightarrow \ \ \ (x - 3)(x - 2y) = 0$
$\Rightarrow \ \ \ x = 3,\ x = 2y$
$\Rightarrow \ \ x = 3,\ \ y = \frac{3}{2}$
$\therefore$ Centre of circle is $\left( 3,\frac{3}{2} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.