Geo4.99

Find the acute angle of intersection of the following circles. $(x-2)^2+y^2=2,(x-2)^2+(y-1)^2=1\,$

From the given circles

$A=(2,0),r_1=\sqrt{2},B=(2,1),r_2=1\,$

$d=AB=\sqrt{(2-2)^2+(0-1)^2}=1\,$

Now the angle between the circles is

$\cos \theta=\frac{1^2-[\sqrt{2}]^2-1^2}{2\cdot\sqrt{2}\cdot1}\,$

$\cos \theta=\frac{-2}{2\sqrt{2}}\,$

$\cos\theta=\frac{-1}{\sqrt{2}}\,$

The acute angle is

$\theta=\frac{\pi}{4}\,$