If $(4, 3)$ and $(- 12, - 1)$ are end points of a diameter of a circle, then the equation of the circle is-

x^{2} + y^{2} - 8 x - 2 y - 51 = 0

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x^{2} + y^{2} + 8 x - 2 y - 51 = 0

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x^{2} + y^{2} + 8 x + 2 y - 51 = 0

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Solution

The correct option is B $x^{2} + y^{2} + 8 x - 2 y - 51 = 0$
End points of diameter are $(4, 3)$ and $(- 12, - 1)$
$∴$ Center of circle $= (\frac{4 - 12}{2}, \frac{3 - 1}{2})$
$= (- 4, 1)$
Radius $= \frac{1}{2} \sqrt{{(4 + 12)}^{2} + {(3 + 1)}^{2}}$
$= \sqrt{68}$
$∴$ Equation of circle is ${(x + 4)}^{2} + {(y - 1)}^{2} = {(\sqrt{68})}^{2}$
$x^{2} + 8 x + 16 + y^{2} - 2 y + 1 = 68$
$x^{2} + y^{2} + 8 x - 2 y - 51 = 0$

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