Geo4.25

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Find the equation of the circle which passes through the points of intersection of x^{2}+y^{2}+3x-4y+5=0\, and x-y+2=0\, and also through the point (2,3).

The formula for equation to the circle is

S+\lambda L=0\, where S=equation of the given circle and L=equation of the line and Lambda is a constant

Hence the equation of the circle is

(x^{2}+y^{2}+3x-4y+5)+\lambda (x-y+2)=0\,

This circle passes through the point (2,3)

Now

(2^{2}+3^{2}+3(2)-4(3)+5)+\lambda (2-3+2)=0\,

18-6+\lambda =0\,

\lambda =-12\,

Hence the equation of the circle is

x^{2}+y^{2}+3x-4y+5+x-y+2=0\,

x^{2}+y^{2}+4x-5y+7=0\,

Main Page:Geometry:Circles