Geo4.89

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Find the equation to the circle whose diameter is the common chord of two circles x^{2}+y^{2}-4x+6y-12=0,x^{2}+y^{2}+2x-2y-23=0\,. Hence find the length of the common chord.

The equation of the common chord of the two circles is

(x^{2}+y^{2}-4x+6y-12)-(x^{2}+y^{2}+2x-2y-23)=0\,

6x-8y-11=0\,

Equation of the circle through the points of intersection of 1 and 2 is

x^{2}+y^{2}-4x+6y-12+k(6x-8y-11)=0\,

x^{2}+y^{2}+x(6k-4)+y(6-8k)-(12+11k)=0\,

Centre of this circle is

[-(3k-2),-(3-4k)]\,

If this circle has the common chord as diameter,the centre of the cirdcle lies on the commn chord, hence

6(2-3k)-8(4k-3)-11=0\,

-50k+25=0,k={\frac  {1}{2}}\,

The equation of the required circle is

2x^{2}+2y^{2}-2x+4y-35=0\,

Radius is {\sqrt  {{\frac  {1}{4}}+1+{\frac  {35}{2}}}}={\frac  {{\sqrt  {75}}}{2}}\,

Length of the common chord of 1 and 2 equations is

2{\frac  {{\sqrt  {75}}}{2}}=5{\sqrt  {3}}\,

Main Page:Geometry:Circles