If the vertices of a triangle are $(2, - 2), (- 1, - 1)$ and $(5, 2)$ , then the equation of its circumcircle is

x^{2} + y^{2} + 3 x + 3 y + 8 = 0

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x^{2} + y^{2} - 3 x - 3 y - 8 = 0

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x^{2} + y^{2} - 3 x + 3 y + 8 = 0

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Solution

The correct option is B $x^{2} + y^{2} - 3 x - 3 y - 8 = 0$
let circumentre be $O (x, y)$

$\begin{matrix} {(O A)}^{2} = {(O B)}^{2} = {(O C)}^{2} {(x - 2)}^{2} + {(y + 2)}^{2} = {(x + 1)}^{2} + {(y + 1)}^{2} - 4 x + 4 y + 8 = 2 x + 2 y + 2 - 6 x + 2 y + 6 = 0 - 3 x + y + 3 = 0....... (1) {(x + 1)}^{2} + {(y + 1)}^{2} = {(x - 5)}^{2} + {(y - 2)}^{2} 2 x + 2 y + 2 = - 10 x - 4 y + 29 12 x + 6 y - 27 = 0 4 x + 2 y - 9 = .0 . . . . . . . . . . (2) - 6 x + 2 y + 6 = 0 4 x + 2 y - 9 = 0 - --------------- - - 10 x + 15 = 0 x = \frac{3}{2} y = 3 \times \frac{3}{2} - 3 {(O A)}^{2} = {(\frac{3}{2} - 2)}^{2} + {(\frac{3}{2} + 2)}^{2} = \frac{1}{4} + \frac{49}{4} = \frac{50}{4} {(x - \frac{3}{2})}^{2} + {(y - \frac{3}{2})}^{2} = \frac{50}{4} x^{2} + y^{2} - 3 x - 3 y = \frac{50}{4} - \frac{18}{4} x^{2} + y^{2} - 3 x - 3 y = 8 \end{matrix}$

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If the vertices of a triangle are (2,−2),(−1,−1) and (5,2), then the equation of its circumcircle is

If the vertices of a triangle are $(2, - 2), (- 1, - 1)$ and $(5, 2)$ , then the equation of its circumcircle is