Centre of circle whose normals 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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None of these

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Solution

The correct option is A $3, \frac{3}{2}$
x2–2xy–3x+6y=0

x(x–3)–2y(x–3)=0

(x–2y)(x–3)=0

x−2y=0 are the normal equations.

WKT normal passes through center intersection of two normal will give center

$x = 3, x = 2 y$

$y = 3 / 2$

Center is $(3, 3 / 2)$

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

The correct option is A 3,32x2–2xy–3x+6y=0x2–2xy–3x+6y=0 x(x–3)–2y(x–3)=0x(x–3)–2y(x–3)=0 (x–2y)(x–3)=0(x–2y)(x–3)=0 x−2y=0x−2y=0 are the normal equations. WKT normal passes through center intersection of two normal will give center x=3,x=2y y=3/2 Center is (3,3/2)

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

The correct option is A $3, \frac{3}{2}$
x2–2xy–3x+6y=0

x(x–3)–2y(x–3)=0

(x–2y)(x–3)=0

x−2y=0 are the normal equations.

WKT normal passes through center intersection of two normal will give center

$x = 3, x = 2 y$

$y = 3 / 2$

Center is $(3, 3 / 2)$