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

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

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

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

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Solution

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

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

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

$⟹$ $x^{2} + 2 x - 4 x - 8 + y^{2} - y + y - 1 = 0$

$⟹$ $x^{2} + y^{2} - 2 x - 9 = 0$

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