Geo4.85

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Show that the circles x^{2}+y^{2}+2ax+c=0,x^{2}+y^{2}+2by+c=0\, touch each other if {\frac  {1}{a^{2}}}+{\frac  {1}{b^{2}}}={\frac  {1}{c^{2}}}\,.

Given circles are

x^{2}+y^{2}+2ax+c=0,x^{2}+y^{2}+2by+c=0\,

The equation of the common tangent of the two circles is

(x^{2}+y^{2}+2ax+c)-(x^{2}+y^{2}+2by+c)=0\,

ax-by=0\,

The peprpendicular distance from (-a,0) to the common tangent =radius {\sqrt  {a^{2}-c}}\,

{\frac  {|-a^{2}|}{{\sqrt  {a^{2}+b^{2}}}}}={\sqrt  {a^{2}-c}}\,

a^{4}=(a^{2}-c)(a^{2}+b^{2})\,

a^{2}b^{2}=a^{2}c+b^{2}c\,

{\frac  {1}{a^{2}}}+{\frac  {1}{b^{2}}}={\frac  {1}{c^{2}}}\,


Main Page:Geometry:Circles