Geo4.75

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Find the equation of the common chord of the circles x^{2}+y^{2}+4x-12y+14=0\, andx^{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}}\,

Main Page:Geometry:Circles