SOLUTION: 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)}}}

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: 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)}}}      Log On


   

Question 1016115: Show that the poles of tangents of the circle %28x-p%29%5E2%2By%5E2=b%5E2 with respect to the circle x%5E2%2By%5E2=a%5E2 lie on the curve %28px-a%5E2%29%5E2=b%5E2%28x%5E2%2By%5E2%29
Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.
Very interesting, but much higher than the school math.

Not for this site.