The angle at which the circles (x−1)2+y2= 10 and x2+(y−2)2=5 intersect is

Discussion Forum : Circles

Question -

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

Options:

A .

\frac{π}{6}

B .

\frac{π}{4}

C .

\frac{π}{3}

D .

\frac{π}{2}

Answer: Option B
:
B

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

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