Geo4.91

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Find the equation of the circle having the common chord of the circles x^{2}+y^{2}+x+5y-2=0,x^{2}+y^{2}+2x+5y-4=0\, as diameter.

Equation of the common chord is

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

-x+2=0,x=2\,

The equation of the circle passing through the point as intersection of the two circles is

x^{2}+y^{2}+x+5y-2+k(x-2)=0\,

Centre of the circle is (-{\frac  {k+1}{2}},{\frac  {-5}{2}})\,

Since the circle has the common chord as diameter,centre lies on the chord.

k+1+4=0,k=-5\,

The required circle is

x^{2}+y^{2}-4x+5y+8=0\,

Main Page:Geometry:Circles