Geo4.86

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Find the length of the common chord of the two circles (x-a)^{2}+y^{2}=a^{2},x^{2}+(y-b)^{2}=b^{2}\,.

Given the circles are

(x-a)^{2}+y^{2}=a^{2},x^{2}+(y-b)^{2}=b^{2}\,

Equation to the common chord of the two equations is

([x-a]^{2}+y^{2}-a^{2})-(x^{2}+(y-b)^{2}-b^{2})=0\,

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

ax-by=0\,

Centre of first circle is (a,0)\, and radius is a.

Length of the perpendicular distance from (a,0) is

{\frac  {a(a)-b(0)}{{\sqrt  {a^{2}+b^{2}}}}}={\frac  {a^{2}}{{\sqrt  {a^{2}+b^{2}}}}}\,

Length of the common chord of the circles is

{\sqrt  {r^{2}-p^{2}}}=2{\sqrt  {[a^{2}-{\frac  {a^{4}}{a^{2}+b^{2}}}]}}={\frac  {2ab}{{\sqrt  {a^{2}+b^{2}}}}}\,


Main Page:Geometry:Circles