Geo4.96

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Find the equation of the circle of radius 2{\sqrt  {2}}\, and passing through the points of intersection of the circles x^{2}+y^{2}-2x-4y+4=0,x^{2}+y^{2}=4\,.

Equation of the common chord of the circles is

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

-2x-4y+8=0\,

x+2y-4=0\,

The equation of circle passing through the intersection of the given circles

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

x^{2}+y^{2}+kx+2ky-(4+4k)=0\,

Given the radius of this circle is 2sqrt2 and the centre is (-k/2,-k).

Now

{\sqrt  {k^{2}+4k^{2}+16+16k}}=4{\sqrt  {2}}\,

5k^{2}+16+16k=32\,

5k^{2}+16k-16=0\,

(5k-4)(k+4)=0\,

k=-4,{\frac  {4}{5}}\,

Therefore,the required circle is

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


Main Page:Geometry:Circles