Find the equation of the circle which passes through the points $(2, - 2)$ and $(3, 4)$ . And whose centre lies on the line $x + y = 2$ .

x^{2} + y^{2} + 7 x + 3 y + 16 = 0

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x^{2} + y^{2} + 7 x - 3 y - 16 = 0

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x^{2} + y^{2} + 4 x + 5 y - 16 = 0

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x^{2} + y^{2} + 5 x + 8 y + 30 = 0

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Solution

The correct option is B $x^{2} + y^{2} + 7 x - 3 y - 16 = 0$
As $(2, - 2)$ passes through the circle.
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0 4 + 4 + 4 g - 4 f + c = 0 4 g + 4 f + 8 + c = 0 ⟶ (1)$
As $(3, 4)$ passes through the circle
$25 + 6 g + 8 f + c = 0 ⟶ (2)$
As center of a circle $(- g, - f)$
$- g - f = 2 g + f = - 2 ⟶ (3) 4 g - 4 f + 8 + c = 0 4 (g - f + 2 + \frac{c}{4}) = 0 g - f + 2 + \frac{c}{4} = 0 - 4 = \frac{c}{4} c = - 16 4 g - 4 f - 8 = 0 \times 2 8 g - 8 f - 16 = 0 6 g + 8 f + 9 = 0 \times 1 6 g + 8 f + 9 = 0 2 g = - 7 g = \frac{- 7}{2} f = \frac{3}{2}$
Equation of circle
$x^{2} + y^{2} + 2 \times \frac{7}{2} x - 2 \times \frac{3}{2} y - 16 = 0$
$x^{2} + y^{2} + 7 x - 3 y - 16 = 0$

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