Question

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

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is B

$\frac{\pi }{4}$

$r_1=\sqrt{10}{r}_2=\sqrt{5}d=\sqrt{(1{)}^2+(2{)}^2}=\sqrt{5}cos\theta =\frac{r_1^2+r_2^2-d^2}{2r_1{r}_2}=\frac{10+5-5}{2\sqrt{10}\sqrt{5}}=\frac{10}{2\times 5\times \sqrt{2}}=\frac{1}{\sqrt{2}}$