Geo4.93

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Find the equation of the circle which passes through the points of intersection of x^{2}+y^{2}-2x+1=0,x^{2}+y^{2}-5x-6y+4=0\, and touch the line 2x-y+3=0\,

Equation of the common chord is

(x^{2}+y^{2}-2x+1)-(x^{2}+y^{2}-5x-6y+4)=0\,

3x+6y-3=0\,

x+2y-1=0\,

Equation of the circle passing through the point of intersection of the two circles is

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

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

Centre of the circle is

({\frac  {2-k}{2}},-k)\,

Radius is

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

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

{\frac  {|2-k+k+3|}{{\sqrt  {5}}}}={\frac  {k{\sqrt  {5}}}{2}}\,

k=\pm 2\,

Hence the equations of the circles are

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


Main Page:Geometry:Circles