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Let $C_1, C_2$ be two concentric circles in the extended complex plane. Is it true that if another "circle" $C$ is orthogonal to both $C_1$ and $C_2$, then $C$ must be a line?

I think that this should be true, because at the point of intersection between $C_1$ and $C$, the tangent of $C$ must intersect the center of $C_1$. Similarly, this must also hold for $C_2$. But since the centers of concentric circles are the same, I think that this might constrain $C$ to taking the form of a line. I'm not sure if this is actually true though, or how one would show the statement.

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Your remark that the tangent of $C$ at the intersection with $C_1$ goes through the center of $C_1$ is the key. Similarly, the tangent of $C_1$ at the intersection with $C$ goes through the center of $C$. If the radius of $C$ is $r$ and the radius of $C_1$ is $r_1$, the distance between the centers is $\sqrt{r^2+r_1^2}$. By the same token, if the radius of $C_2$ is $r_2$ the distance between the centers of $C_2$ and $C$ is $\sqrt{r^2+r_2^2}$ These disagree unless $r$ is infinite.

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(This probably should be a comment)

Here's one way to go about it: construct two concentric circles of arbitrary radius: $x^2+y^2=r_1^2$ and $x^2+y^2=r_2^2$; and a third circle of arbitrary radius that is orthogonal to one of the circles. Now, try to prove that the tangents at the intersection points of the other concentric circle and the circle orthogonal to the first circle cannot be perpendicular...

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Thanks for the tip! I reasoned out Ross's argument using what you said, but since he already posted the answer, I'll just accept it. –  user1736 Feb 14 '11 at 4:55
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