Geo4.52

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Show that the poles of tangents of the circle (x-p)^{2}+y^{2}=b^{2}\, with respect to the circle x^{2}+y^{2}=a^{2}\, lie on the curve (px-a^{2})^{2}=b^{2}(x^{2}+y^{2})\,

Let (x1,y1) be the pole w.r.t the circle x^{2}+y^{2}=a^{2}\,

Equation of the polar is xx_{1}+yy_{1}-a^{2}=0\,

Given that this equation is a tangent to the circle (x-p)^{2}+y^{2}=b^{2}\,

Hence

|{\frac  {px_{1}+0-a^{2}}{{\sqrt  {x_{1}^{{2}}+y_{1}^{{2}}}}}}|=b\,

(px_{1}-a^{2})^{2}=b^{2}(x_{1}^{{2}}+y_{1}^{{2}})\,

Therefore the locus of tangents o fthe given circle lie on the curve

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


Main Page:Geometry:Circles