# Geo4.43

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

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