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^{2}_{1} + 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}$