The equation of the circle having $(4, - 1)$ and $(2, - 1)$ as the endpoints of its diameter is

$x^{2} + y^{2} - 6 x + 2 y - 9 = 0$

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$x^{2} + y^{2} - x + 2 y + 9 = 0$

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$x^{2} + y^{2} + x + y - 9 = 0$

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$x^{2} + y^{2} - 6 x + 2 y + 9 = 0$

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Solution

The correct option is D
$x^{2} + y^{2} - 6 x + 2 y + 9 = 0$

Equation of a circle when the end points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ of its diameter are given, is given by,

$(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0$

So, when $(4, - 1)$ and $(2, - 1)$ are the end points of the diameter, equation of the circle would become:

$(x - 4) (x - 2) + (y + 1) (y + 1) = 0$

$x^{2} - 6 x + 8 + y^{2} + 2 y + 1 = 0$

$x^{2} + y^{2} - 6 x + 2 y + 9 = 0$

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