Geo4.44

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Find the equation to the circle passing through the points of intersection of the lines x+2y-1=0 and the circle x^{2}+y^{2}-2x+1=0\, and touching the line 2x-y+3=0.

The equation of the circle is

x^{2}+y^{2}-2x+1+k(x+2y-1)=0\,

x^{2}+y^{2}+x(k-2)+2ky+1-k=0\,

The centre of the circle is ({\frac  {2-k}{2}},-k)\, and radius is {\sqrt  {{\frac  {(k-2)^{2}}{4}}+k^{2}+k-1}}\,

This circle touches the line 2x-y+3=0\,

Hence the distance fromt the centre =radius

|{\frac  {2-k+k+3}{{\sqrt  {5}}}}|={\sqrt  {{\frac  {(k-2)^{2}}{4}}+k^{2}+k-1}}\,

20=k^{2}+4-4k+4k^{2}+4k-4\,

5k^{2}=20\,

k=\pm 2\,

Hence the equations of the circle

x^{2}+y^{2}+4y-1=0,x^{2}+y^{2}-4x-4y+3=0\,

Main Page:Geometry:Circles