Centre of circle whose normals are x^2- 2xy - 3x + 6y = 0, is: - Vidyadip

MCQ

Centre of circle whose normals are $x^2- 2xy - 3x + 6y = 0$, is:

✓
$\big(3, \frac{3}{2}\big)$
B
$\big(3, -\frac{3}{2}\big)$
C
$\big(\frac{3}{2},3\big)$
D
None of these

✓

Answer

Correct option: A.

$\big(3, \frac{3}{2}\big)$

$x^2- 2xy - 3x + 6y = 0$
$\Rightarrow (x - 3) (x - 2y) = 0$
$\Rightarrow x = 3$ and $x = 2y$ are two normals.
The intersection point of these two normals will be the centre of the circle.
$\therefore$ for $x = 3$
$\Rightarrow\text{y}=\frac{\text{x}}{2} = \frac{3}{2}$
The intersection point is $\big(3, \frac{3}{2}\big)$ the centre of the given circle is $\big(3, \frac{3}{2}\big)$

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