SOLUTION: Show that the locus of the poles of tangents to the circle {{{x^2+y^2=a^2}}} with respect to the circle {{{(x+a)^2+y^2=2a^2}}} is {{{y^2+4ax=0}}}

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Show that the locus of the poles of tangents to the circle {{{x^2+y^2=a^2}}} with respect to the circle {{{(x+a)^2+y^2=2a^2}}} is {{{y^2+4ax=0}}}      Log On


   

Question 1016117: Show that the locus of the poles of tangents to the circle x%5E2%2By%5E2=a%5E2 with respect to the circle %28x%2Ba%29%5E2%2By%5E2=2a%5E2 is y%5E2%2B4ax=0
Answer by ikleyn(52800) About Me  (Show Source):
You can put this solution on YOUR website!
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Very interesting, but much higher than the school math.

Not for this site.

Beside of it, the words 

 ". . . the poles of tangents to the circle . . .  with respect to the circle . . . " require additional explanations.

Not all of us are familiar with this terminology.