Geo4.92

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Show that the length of the common chord of the two circles x^{2}+y^{2}+2gx+c=0\, and x^{2}+y^{2}+2fy-c=0\, is 2{\sqrt  {{\frac  {(g^{2}-c)(f^{2}+c)}{g^{2}+f^{2}}}}}\,

Equation of the common chord of the two circles is

(x^{2}+y^{2}+2gx+c)-(x^{2}+y^{2}+2fy-c)=0\,

2gx-2fy+2c=0\,

gx-fy+c=0\,

Centre is (-g,0) and Radius of the first circle is

r={\sqrt  {g^{2}-c}}\,

The perpendicular distance from the centre to the common chord is

p={\frac  {-g^{2}+c}{{\sqrt  {g^{2}+f^{2}}}}}\,

The length of the chord is

2{\sqrt  {r^{2}-p^{2}}}=2{\sqrt  {{\frac  {(g^{2}-c)(g^{2}+f^{2})-(g^{2}-c)^{2}}{g^{2}+f^{2}}}}}\,

2{\sqrt  {{\frac  {(g^{2}-c)(g^{2}+f^{2}-g^{2}+c)}{g^{2}+f^{2}}}}}\,

2{\sqrt  {{\frac  {(g^{2}-c)(f^{2}+c)}{g^{2}+f^{2}}}}}\,


Main Page:Geometry:Circles