Geo4.78

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If the two circles x^{2}+y^{2}+2gx+2fy=0,x^{2}+y^{2}+2g_{1}x+2f_{1}y=0\, touch eachother,prove that fg_{1}=f_{1}g\,

Centre of the given circles are

A(-g,-f),B(-g_{1},-f_{1})\,

Radii of the given circles are

r_{1}={\sqrt  {g^{2}+f^{2}}},r_{2}={\sqrt  {g_{1}^{{2}}+f_{1}^{{2}}}}\,

Since both of the equations don't contain the constants,the two equations touch eachother at O(0,0).

The two circles touch each other.A,O,B are collinear.

Slope of AO is equal to slope of BO

{\frac  {g}{g_{1}}}={\frac  {f}{f_{1}}}\,

gf_{1}=fg_{1}\,

This is the required condition.


Main Page:Geometry:Circles