Explanation:
D Given,
\(x^2+y^2-2 x-2 y=0\)
\(x^2+y^2=4\)
\(\text { Now, } c_1(1,1), r_1=\sqrt{1^2+1^2}=\sqrt{2}\)
\(C_2(0,0), r_2=2\)
\(\text { If } \theta \text { is the angle of intersection then }\)
\(\cos \theta=\frac{r_1^2+r_2^2-\left(c_1 c_2\right)^2}{2 r_1 r_2}\)
\(=\frac{2+4-(\sqrt{2})^2}{2 \cdot \sqrt{2} \cdot 2}=\frac{1}{\sqrt{2}} \Rightarrow \theta=45^{\circ}\)
Ans: d
Exp:D Given,
Now, \(\mathrm{C}_1(1,1), \mathrm{r}_1=\sqrt{1^2+1^2}=\sqrt{2}\)
If \(\theta\) is the angle of intersection then