Geo4.47

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Find the pole of the line 3x+4y-45=0 with respect to the circle x^{2}+y^{2}-6x-8y+5=0\,

Let (x1,y1) be the pole of the given line with respect to the given circle.

Therefore,the equation of the polar is

xx_{1}+yy_{1}-3(x+x_{1})-4(y+y_{1})+5=0\,

(x_{1}-3)x+(y_{1}-4)y-(3x_{1}+4y_{1}-5)=0\,

This equation and the given line equation represent the same line,

Hence

{\frac  {x_{1}-3}{3}}={\frac  {y_{1}-4}{4}}={\frac  {3x_{1}+4y_{1}-5}{45}}\,

By ratio and proportion,each is equal to

{\frac  {3(x_{1}-3)+4(y_{1}-4)-(3x_{1}+4y_{1}-5)}{3(3)+4(4)-45}}={\frac  {20}{20}}=1\,

Hence

{\frac  {x_{1}-3}{3}}=1,x_{1}=6\,

{\frac  {y_{1}-4}{4}}=1,y_{1}=8\,

Hence the pole of the line with respect to the circle is (6,8).


Main Page:Geometry:Circles