Geo4.90

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

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