# 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.

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?

• 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.
– Jens
Commented Oct 17, 2018 at 16:34
• An example would be if $O_1=(0,0)$, $O_2=(1,0)$, $r_1=1$, $r_2=3$ and $k=0.5$.
– Jens
Commented Oct 17, 2018 at 17:23

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.

• Wouldn't it only be an Apollonius circle if $\frac {\sqrt {PO_1^2}}{\sqrt {PO_2^2}} = m$, for some constant $m$?
– Jens
Commented Oct 17, 2018 at 16:24
• @Jens Apollonius circle is built from two foci. They are the points of contacts of the tangents from P to the two circles.
– Mick
Commented Oct 17, 2018 at 17:36
• 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.
– Jens
Commented Oct 17, 2018 at 17:39
• @Jens See picture added. Or draw one to verify that.
– Mick
Commented Oct 17, 2018 at 17:43
• OK, but it seems to me the two foci wouldn't remain fixed points. Isn't that a pre-requisite?
– Jens
Commented Oct 17, 2018 at 17:51