Geo4.75

Find the equation of the common chord of the circles $x^{2}+y^{2}+4x-12y+14=0\,$ and$x^{2}+y^{2}-14x+6y-22=0\,$. Find the points of intersection of the circles.Also find the length of the common chord.

Given circles are

$x^{2}+y^{2}+4x-12y+14=0,x^{2}+y^{2}-14x+6y-22=0\,$

Common chord is

$S_{1}-S_{2}=0\,$

$(x^{2}+y^{2}+4x-12y+14)-(x^{2}+y^{2}-14x+6y-22)=0\,$

$18x-18y+36=0\,$

$x-y+2=0\,$

To find the points of intersection,we can simplify the chord equation with any of the circle equations.

Now $y=x+2\,$

Substituting this in the first circle equation,

$x^{2}+(x+2)^{2}+4x-12(x+2)+14=0\,$

$2x^{2}+4x+4+4x-12x-24+14=0\,$

$2x^{2}-4x-6=0\,$

$x^{2}-2x-3=0\,$

$(x-3)(x+1)=0,x=3,-1\,$

The values of y are

$y=3+2,-1+2\,$

hence the points are $(3,5),(-1,1)\,$

Length of the common chord =Distance between the points of intersection.

Hence length is

${\sqrt {(3+1)^{2}+(5-1)^{2}}}={\sqrt {32}}=4{\sqrt {2}}\,$