Geo4.26

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Find the equation of the circle which passes through the points of intersection of x^{2}+y^{2}-8x-2y+7=0\, and 4x+12y+1=0\, and which has its centre on y-axis.

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}-8x-2y+7)+\lambda (4x+12y+1)=0\,

The circle has its centre on y axis means x=0.

Now

Centre of the above equation is

x^{2}+y^{2}-8x+4\lambda x-2y+12\lambda y+7+\lambda =0\,

x^{2}+y^{2}-2x(4-2\lambda )-2y(1-6\lambda )+7+\lambda =0\,

Now

-g=-(-(4-2\lambda ))=0\,

\lambda =2\,

Hence the equation of the circle is

x^{2}+y^{2}-8x-2y+7+2(4x+12y+1)=0\,

x^{2}+y^{2}-8x-2y+7+8x+24y+2=0\,

x^{2}+y^{2}+22y+9=0\,

Main Page:Geometry:Circles