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

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Aicharya · Accepted Answer

Here circles are $ \begin{array}{l} x^{2}+y^{2}-2 x-2 y=0...(1) \ x^{2}+y^{2}=4...(2) \ 	ext { Now, } \mathrm{C}_{1}(1,1), r_{1}=\sqrt{1^{2}+1^{2}}=\sqrt{2} \ C_{2}(0,0), r_{2}=2 \end{array} $ If $	heta$ is the angle of intersection then $ \begin{array}{l} \cos 	heta=\frac{r_{1}^{2}+r_{2}^{2}-\left(c_{1} c_{2}ight)^{2}}{2 r_{1} r_{2}} \ =\frac{2+4-(\sqrt{2})^{2}}{2 \cdot \sqrt{2.2}}=\frac{1}{\sqrt{2}} \Rightarrow 	heta=45^{\circ} \end{array} $

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

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