The equation of the circle whose one diameter is $P Q$ , where the ordinates of $P, Q$ are the roots of the equation $x^{2} + 2 x - 3 = 0$ and the abscissae are the roots of the equation $y^{2} + 4 y - 12 = 0$ , is

x^{2} + y^{2} + 2 x + 4 y - 15 = 0

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x^{2} + y^{2} - 4 x - 2 y - 15 = 0

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x^{2} + y^{2} + 4 x + 2 y - 15 = 0

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Solution

The correct option is B $x^{2} + y^{2} + 2 x + 4 y - 15 = 0$
Equation of circle with PQ as diameter is given by,
$(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0$
$\Rightarrow x^{2} - (x_{1} + x_{2}) x + x_{1} x_{2} + y^{2} - (y_{1} + y_{2}) y + y_{1} y_{2} = 0$
$\Rightarrow x^{2} + y^{2} + 2 x + 4 y - 15 = 0$
Where $x_{1}, x_{2}, y_{1}, y_{2}$ are roots of the given quadratics.

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The equation of the circle whose one diameter is PQ, where the ordinates of P,Q are the roots of the equation x2+2x−3=0 and the abscissae are the roots of the equation y2+4y−12=0, is

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The equation of the circle whose one diameter is $P Q$ , where the ordinates of $P, Q$ are the roots of the equation $x^{2} + 2 x - 3 = 0$ and the abscissae are the roots of the equation $y^{2} + 4 y - 12 = 0$ , is