Geo4.121

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Find the equation to the circle touching the line 2x-y+3=0\, and belonging to the coaxal system determined by x^{2}+y^{2}-2x+1=0\, and the radical axis x+2y-1=0\,

The equation of the circle belonging to the coaxal sysem detemined by

x^{2}+y^{2}-2x+1=0\, and the radical axis x+2y-1=0\, is

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

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

centre of the circle is (1-{\frac  {\lambda }{2}},-\lambda )\,

Radius is {\sqrt  {[{\frac  {2-\lambda }{2}}]^{2}+\lambda ^{2}-(1-\lambda )}}={\frac  {\lambda {\sqrt  {5}}}{2}}\,

Since the circle touches the line 2x-y+3=0,we have

{\frac  {|2-\lambda +\lambda +3|}{{\sqrt  {4+1}}}}={\frac  {\lambda {\sqrt  {5}}}{2}}\,

By solving,

\lambda =\pm 2\,

Hence the equation of the required circle is

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


Main Page:Geometry:Circles