Question

The angle of intersection of the two circles $x^2+y^2-2x-2y=0$ and $x^2+y^2=4$, is

• A. $30^{\circ }$
• B. $60^{\circ }$
• C. $90^{\circ }$
• D. $45^{\circ }$

Answer 1

Answer: (D)

Here circles are
$x^2+y^2-2x-2y=0$ ...(1)
$x^2+y^2=4$ ...(2)
Now, $C_1(1,1),r_1=\sqrt{1^2+1^2}=\sqrt{2}$
$C_2(0,0),r_2=2$
If $\theta$ is the angle of intersection then
$\cos \theta =\frac{r_1^2+r_2^2-{\left(c_1{c}_2\right)}^2}{2r_1{r}_2}$
$=\frac{2+4-{\left(\sqrt{2}\right)}^2}{2\sqrt{2}.2.}=\frac{1}{\sqrt{2}}$
$\Rightarrow \theta =45^{\circ }$