Geo4.99

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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}}\cdot 1}}\,

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

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

The acute angle is

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

Main Page:Geometry:Circles