The angle at which the circles $(x - 1)^{2} + y^{2} = 10$ and $x^{2} + (y - 2)^{2} = 5$ 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 C $\frac{π}{4}$
From given $r_{1} = \sqrt{10}, C_{1} (1, 0)$ and

$r_{2} = \sqrt{5}, C_{2} (0, 2)$
$d = C_{1} C_{2} = \sqrt{5}$
If $θ$ is the angle between the circle, then
$cos θ = \frac{| d^{2} - r_{1}^{2} - r_{2}^{2} |}{2 r_{1} r_{2}}$
$= \frac{| 5 - 10 - 5 |}{2 \sqrt{10} \sqrt{5}}$
$= \frac{1}{\sqrt{2}}$
Hence, $θ = \frac{π}{4}$

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