Find the equation of the circle when the end points of a diameter are $A (- 2, 3)$ and $B (3, - 5)$

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Solution

Using diameter form,the equation of the circle having $A (- 2, 3)$ and $B (3, - 5)$ as the end points of a diameter is
$(x - (- 2)) (x - 3) + (y - 3) (y - (- 5)) = 0$
$\Rightarrow (x + 2) (x - 3) + (y - 3) (y + 5) = 0$
$\Rightarrow x^{2} + 2 x - 3 x - 6 + y^{2} - 3 y + 5 y - 15 = 0$
$\Rightarrow x^{2} + y^{2} - x + 2 y - 21 = 0$ is the required equation of the circle.

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