Geo4.90

From Example Problems
Jump to: navigation, search

Find the equation of the circle described on the common chord of the circles x^{2}+y^{2}-4x-6y-12=0,x^{2}+y^{2}+6x+4y-12=0\, as diameter.

The equation of the common chord of the circles is

(x^{2}+y^{2}-4x-6y-12)-(x^{2}+y^{2}+6x+4y-12)\,.

-10x-10y=0\,

x-y=0\,

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

x^{2}+y^{2}-4x-6y-12+k(x-y)=0\,

x^{2}+y^{2}-x(4-k)-y(6+k)-12=0\,

Centre of the circle is

{\frac  {4-k}{2}},{\frac  {6+k}{2}}\,

If the circle has the common chord as diameter,the point must lie on the chord.

4-k-6-k=0,k=-1\,

Hence the equation of the circle is

x^{2}+y^{2}-5x-5y-12=0\,

Main Page:Geometry:Circles