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Consider two circles with radii $a$ and $b$ and centers $(a, 0)$ and $(b, 0)$ respectively with $0 < a < b$.

Let $c$ be the center of any circle in the crescent shaped region M between the two circles and tangent to both (See figure below).

figure

Determine the locus of c as its circle traverses through region M maintaining tangency.

Attempt: Let $y=k(x-a)$ be a line passing through $(a,0)$ and intersecting the two circles.

The equation of the two cirlces is,

$$\large (x-a)^2 + y^2 = a^2$$ $$\large (x-b)^2 + y^2 = b^2$$

Now $c$ clearly is the mid-point of the points of intersection of the assumed line with the circles (as it must be equidistant from both the inner and outer circles). Thus substituting to find the midpoint,

$$\large \implies (x-a)^2 + k^2(x-a)^2 = a^2.......(i)$$

and,

$$\large \implies (x-b)^2 + k^2(x-a)^2 = b^2.......(ii)$$

The first equation can be solved by,

$$\large (x-a)^2(1+k^2)=a^2$$ $$\large (x-a)^2=\frac{a^2}{1+k^2}$$

Implying, $$x=a \pm \sqrt{\frac {a^2}{1+k^2}}, y=\pm k\sqrt{\frac {a^2}{1+k^2}}$$.

How do I go about solving $(ii)$ similarly. Also any shorter methods?

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Let $A$,$B$, and $C$ be the centers of the circles $(a)$,$(b)$, and $(c)$, respectively. We have $CA + CB = a + b$. Therefore, $C$ is on an ellipse with focal points $(a,0)$ and $(b,0)$ that passes through the origin. It is straightforward to verify that every point on that ellipse you also find a corresponding circle tangent to $(a)$ and $(b)$.

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  • $\begingroup$ You could probably elaborate on the line $CA+CB=a+b$ that took me a while to get. Good solution by the way. Thanks! $\endgroup$ – Guy Feb 25 '14 at 16:04

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