The circles $x^{2} + y^{2} + x + y = 0$ and $x^{2} + y^{2} + x - y = 0$ intersect at an angle of

π / 6

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π / 4

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π / 3

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π / 2

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Solution

The correct option is D $π / 2$
$x^{2} + y^{2} + x + y = 0$

$C_{1} = (\frac{- 1}{2}, \frac{1}{2}), r_{1} = \sqrt{\frac{1}{2}}$

$x^{2} + y^{2} + x - y = 0$

$C_{2} = (\frac{- 1}{2}, \frac{1}{2}), r_{2} = \sqrt{\frac{1}{2}}$

If $θ$ is the angle between circles

$cos θ = \frac{C_{1} C_{2}^{2} - r_{1}^{2} - r_{2}^{2}}{2 r_{1} r_{2}}$

$= \frac{0^{2} + 1^{2} - \frac{1}{2} - \frac{1}{2}}{2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}}$

$cos θ = 0$

$θ = \frac{π}{2}$

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