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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

$$(x_1−p)^2+y_1^2=b^2 \tag{2}$$

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