Geo4.87

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Find the length of the common chord of the two circles x^{2}+y^{2}+4x-12y+14=0,x^{2}+y^{2}-14x+6y-22=0\,

Given circles are

x^{2}+y^{2}+4x-12y+14=0,x^{2}+y^{2}-14x+6y-22=0\,

Equation to the common chord is

(x^{2}+y^{2}+4x-12y+14)-(x^{2}+y^{2}-14x+6y-22)=0\,

18x-18y+36=0\,

x-y+2=0\,

Centre of the first circle is (-2,6),radius {\sqrt  {4+36-14}}={\sqrt  {26}}\,.

the length of the perpendicular is

p={\frac  {-2-(-6)+2}{{\sqrt  {1+1}}}}={\frac  {6}{{\sqrt  {2}}}}\,

Length of the common chord is

2{\sqrt  {r^{2}-p^{2}}}=2{\sqrt  {26-{\frac  {36}{2}}}}=2{\sqrt  {8}}=4{\sqrt  {2}}\,


Main Page:Geometry:Circles