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Let $\omega_1$ and $\omega_2$ be circles with respective centres $O_1$ and $O_2$ and respective radii $r_1$ and $r_2$. Let $k\in\mathbb{R}\setminus\{1\}$. Prove that the set of points $P$ such that $$PO_1^2-r_1^2=k(PO_2^2-r_2^2)$$ is a circle.

It is clear to me how to do this algebraically. If I let $P$ be $(x,y)$, assume without loss of generality that $O_1$ is $(0,0)$ and let $O_2=(a,b)$, then $$x^2+y^2-r_1^2=k((x-a)^2+(y-b)^2-r_2^2),$$ so $$(k-1)x^2+(k-1)y^2-2kax-2kby+ka^2+kb^2-kr_2^2+r_1^2=0$$ and so on.

But is there an easy way of proving this geometrically?

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    $\begingroup$ The statement isn't true without some caveats. There are combinations of $O_1$, $r_1$, $O_2$, $r_2$ and $k$ where the resulting "circle" has an imaginary radius. $\endgroup$
    – Jens
    Commented Oct 17, 2018 at 16:34
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    $\begingroup$ An example would be if $O_1=(0,0)$, $O_2=(1,0)$, $r_1=1$, $r_2=3$ and $k=0.5$. $\endgroup$
    – Jens
    Commented Oct 17, 2018 at 17:23

1 Answer 1

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Note that the given condition is equivalent to $\dfrac {\sqrt {PO_1^2-r_1^2}}{\sqrt {(PO_2^2-r_2^2)}} = \sqrt {k}$, a constant.

This satisfies the condition of a circle, the Apollonius circle.

Picture added:-

enter image description here

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  • $\begingroup$ Wouldn't it only be an Apollonius circle if $\frac {\sqrt {PO_1^2}}{\sqrt {PO_2^2}} = m$, for some constant $m$? $\endgroup$
    – Jens
    Commented Oct 17, 2018 at 16:24
  • $\begingroup$ @Jens Apollonius circle is built from two foci. They are the points of contacts of the tangents from P to the two circles. $\endgroup$
    – Mick
    Commented Oct 17, 2018 at 17:36
  • $\begingroup$ I understand that. But the ratio of the distances from those foci must be a constant. When you include $r_1$ and $r_2$ in the equation, this doesn't seem to be true. $\endgroup$
    – Jens
    Commented Oct 17, 2018 at 17:39
  • $\begingroup$ @Jens See picture added. Or draw one to verify that. $\endgroup$
    – Mick
    Commented Oct 17, 2018 at 17:43
  • $\begingroup$ OK, but it seems to me the two foci wouldn't remain fixed points. Isn't that a pre-requisite? $\endgroup$
    – Jens
    Commented Oct 17, 2018 at 17:51

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