The equation of the circle having $(4, 2)$ and $(2, - 12)$ as the end points of its diameter is

$x^{2} + y^{2} + 6 x + 10 y + 16 = 0$

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$x^{2} + y^{2} - 6 x + 10 y - 16 = 0$

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$x^{2} + y^{2} + 10 x - 6 y + 16 = 0$

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$x^{2} + y^{2} - 10 x + 6 y - 16 = 0$

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

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Solution

The correct option is B
$x^{2} + y^{2} - 6 x + 10 y - 16 = 0$

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 $(4, 2)$ and $(2, - 12)$ are the end points of the diameter, equation of the circle would become:

$(x - 4) (x - 2) + (y - 2) (y + 12) = 0$

$x^{2} - 2 x - 4 x + 8 + y^{2} + 12 y - 2 y - 24 = 0$

$x^{2} + y^{2} - 6 x + 10 y - 16 = 0$

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