Geo4.43

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Find the equation to the circle passing through the points of intersection of the lines x+2y-4=0 and the circle x^{2}+y^{2}=4\, and touching the line x+2y=5.

Equation to any circle throough the intersection of the given circle is

x^{2}+y^{2}-4+k(x+2y-4)=0\,

x^{2}+y^{2}+kx+2ky-4k-4=0\,

Centre of the circle is (\frac{-k}{2},-k) and radius is {\sqrt  {{\frac  {k^{2}}{4}}+k^{2}+4k+4}}\,

The circle touches the line x+2y-5=0\,

Hence the distance from the centre=radius

|{\frac  {{\frac  {-k}{2}}-2k-5}{{\sqrt  {5}}}}|={\sqrt  {{\frac  {k^{2}}{4}}+k^{2}+4k+4}}\,

({\frac  {5k}{2}}+5)^{2}={\frac  {5}{4}}(5k^{2}+16k+16)\,

{\frac  {25}{4}}(k+2)^{2}={\frac  {5}{4}}(5k^{2}+16k+16)\,

Simplifying and solving we get k=-1

The required circle is

x^{2}+y^{2}-x-2y=0\,

Main Page:Geometry:Circles