The angle at which the circles $(x - 1)^{2} + y^{2} = 10 a n d x^{2} + (y - 2)^{2} = 5$ intersect is

$\frac{π}{6}$

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$\frac{π}{4}$

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$\frac{π}{3}$

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$\frac{π}{2}$

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Solution

The correct option is B
$\frac{π}{4}$

$r_{1} = \sqrt{10} r_{2} = \sqrt{5} d = \sqrt{(1)^{2} + (2)^{2}} = \sqrt{5} c o s θ = \frac{r_{1}^{2} + r_{2}^{2} - d^{2}}{2 r_{1} r_{2}} = \frac{10 + 5 - 5}{2 \sqrt{10} \sqrt{5}} = \frac{10}{2 \times 5 \times \sqrt{2}} = \frac{1}{\sqrt{2}}$

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