Geo4.57

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Find the equation of the chord of the circle x^{2}+y^{2}+4x-2y-20=0\,having ({\frac  {1}{2}},{\frac  {-7}{2}})\, as its midpoint.

From the given circle,

g=2,f=-1,c=-20\,

The equation of the chord whose midpoint is {\frac  {1}{2}},{\frac  {-7}{2}}\,

S_{1}=S_{{11}}\,

Now the required equation is

x({\frac  {1}{2}})+y({\frac  {-7}{2}})+2(x+{\frac  {1}{2}})-1(y-{\frac  {7}{2}})-20={\frac  {1}{4}}+{\frac  {49}{4}}+4({\frac  {1}{2}})+2({\frac  {7}{2}})-20\,

x-7y+4x+2-2y+7-40={\frac  {1+49+8+28-80}{2}}\,

2x-14y+8x-62=6\,

2x-14y-68=0\,

x-7y-34=0\,

Main Page:Geometry:Circles