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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 $(px-a^2)^2 = b^2(x^2+y^2)$.

My attempt:

Tangent to $(x−p)^2+y^2=b^2$ at $(x_1,y_1)$ is

$$(x_1−p)(x−p)+yy_1=b^2$$

$$(x_1−p)x+yy_1 = b^2−p^2+px_1 \tag{1}$$

with

$$(x_1−p)^2+y_1^2=b^2 \tag{2}$$

  • 0
    Related? http://math.stackexchange.com/questions/1412476/prove-that-the-locus-of-the-poles-of-tangents-to-the-parabola-y2-4ax-with-res – 2017-01-07
  • 0
    Answer http://exampleproblems.com/wiki/index.php/Geo4.52 – 2017-01-07

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