Geo4.88

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Prove that the length of the common chord of the circles (x-a)^{2}+(y-b)^{2}=c^{2},(x-b)^{2}+(y-a)^{2}=c^{2}\, is {\sqrt  {[4c^{2}-2(a-b)^{2}]}}\,. Hence find the condition that the circles may touch.

The equation of the common chord of the two circles is

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

(x^{2}+y^{2}-2ax-2by+a^{2}+b^{2}-c^{2})-(x^{2}+y^{2}-2bx-2ay+a^{2}+b^{2}-c^{2})\,

2(b-a)x-2(b-a)y=0\,

x-y=0\,

Centre of the first circle is (a,b) and radius is c.

p={\frac  {a-b}{{\sqrt  {1+1}}}}={\frac  {a-b}{{\sqrt  {2}}}}\,

Length of the common chord is

2{\sqrt  {r^{2}-p^{2}}}=2{\sqrt  {[c^{2}-{\frac  {(a-b)^{2}}{2}}]}}={\sqrt  {[4c^{2}-2(a-b)^{2}]}}\,

The circles touch eachother if the length of the common chord is zero.

4c^{2}-2(a-b)^{2}=0\,

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

Main Page:Geometry:Circles