The pair of straight lines $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ meet the coordinate axes in concyclic points. The equation of the circle through those concyclic points is

a x^{2} + a y^{2} + 2 g x + 2 f y + c = 0

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x^{2} + y^{2} + 2 g x - 2 f y + c = 0

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x^{2} + y^{2} - 2 g x - 2 f y - c = 0

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a x^{2} + a y^{2} - 2 g x - 2 f y - c = 0

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Solution

The correct option is C $a x^{2} + a y^{2} + 2 g x + 2 f y + c = 0$
For a curve of the form $a x^{2} + b y^{2} + 2 h x y + 2 g x + 2 f y + c = 0$ is a circle if $h = 0, a = b, g^{2} + f^{2} - c > 0$
Hence option a

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Similar questions

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

a x^{2} - 2 h x y + b y^{2} + 2 g x + 2 f y + c + \frac{4 f g}{c} x y = 0

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

a^{'} x^{2} + 2 h^{'} x y + b^{'} y^{2} + 2 g^{'} x + 2 f y + c^{'} = 0

Which of the following equations represents a circle with centre $(g, f)$ and radius $\sqrt{g^{2} + f^{2} + c}$ ?

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

a^{'} x^{2} + 2 h^{'} x y + b^{'} y^{2} + 2 g^{'} x + 2 f^{'} y + c^{'} = 0

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