Centre of circle whose normal's are $x^{2} - 2 x y - 3 x + 6 y = 0$ , is

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

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(3, - \frac{3}{2})

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

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Solution

The correct option is A $(3, \frac{3}{2})$
$x^{2} - 2 x y - 3 x + 6 y = 0$

$\Rightarrow (x - 3) (x - 2 y) = 0$

Hence, $x = 3$ and $x = 2 y$ are two normals.

The intersection point of these two normals will be the centre of the circle.
$∴$ for $x = 3$ ,

$\Rightarrow$ $y = \frac{x}{2} = \frac{3}{2}$

Hence, the intersection point is $(3, \frac{3}{2})$ .
Hence, the centre of the given circle is $(3, \frac{3}{2})$ .

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Centre of circle whose normal's are x2−2xy−3x+6y=0, is

The correct option is A (3, 32)x2−2xy−3x+6y=0⇒(x−3)(x−2y)=0Hence, x=3 and x=2y are two normals.The intersection point of these two normals will be the centre of the circle.∴ for x=3,⇒y=x2=32Hence, the intersection point is (3,32).Hence, the centre of the given circle is (3,32).

Centre of circle whose normal's are $x^{2} - 2 x y - 3 x + 6 y = 0$ , is