# Construction of touching circle

My question is:

Consider five collinear points $D$, $A$, $C$, $B$ and $E$ such that $DA=AC=a$ , $CB=BE=b$.
Let M be the midpoint of $DE$. Let $S_1$ be a circle with center $A$ and radius $a$, Let $S_2$ be a circle with center $B$ and radius $b$, Let $S_3$ be a circle with center $M$ and radius $a + b$.
Construct circle $S$ using straightedge and compass such that it touches the three circles $S_1$, $S_2$ and $S_3$.

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By "construction", do you mean compass and straightedge or simply finding the center and radius? – Karolis Juodelė Jun 14 '12 at 16:03
@KarolisJuodelė i mean compass and straightedge – mgh Jun 14 '12 at 16:05
Have you tried finding the center and radius algebraically? – Karolis Juodelė Jun 14 '12 at 18:50
@KarolisJuodelė after you asked i tried it algebraically but could not get it that way please will you tell how to get it algebraically – mgh Jun 15 '12 at 6:50

Drop perpendiculars to $A$ and $B$ as in the diagram below and let $F$ and $G$ be the respective intersections with $S_1$ and $S_2$.

$\hspace{3cm}$

Draw $S_4$ centered at $F$ and passing through $C$ and $S_5$ centered at $G$ also passing through $C$. $S_4$ intersects $S_2$ at $C$ and $J$, and $S_5$ intersects $S_1$ at $C$ and $H$. Both $S_4$ and $S_5$ intersect $S_3$ at $K$.

The circle tangent to $S_1$, $S_2$, and $S_3$ passes through $H$, $J$, and $K$. The center of this circle can be found at the intersection of the perpendicular bisectors of $\overline{HJ}$, $\overline{JK}$, and $\overline{HK}$.

Justification of the Construction:

The construction was gotten by inverting the diagram above through a circle centered at $C$. Circles $S_1$ and $S_2$ become parallel lines passing through the images of $D$ and $E$ with the image of $S_3$ tangent to them.

$\hspace{5cm}$

A circle tangent to all three of these is the blue circle above the image of $S_3$ also tangent to the images of $S_1$ and $S_2$.

The lines which are the images of $S_4$ and $S_5$ are perpendicular, so that $S_4$ and $S_5$ must intersect perpendicularly at $K$ and $C$. Thus, $F$ and $G$ were placed at the tops of $S_1$ and $S_2$ so that $\triangle DFC$ and $\triangle CGE$ are $45^\circ{-}90^\circ{-}45^\circ$ triangles.

The duality between the diagrams justifies the rest.

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Very nice. Thanks for the ping! Step two would be to give this construction without using straightedge :) – t.b. Jun 15 '12 at 5:35
@robjohn thanks...explained in an excellent way ._. – mgh Jun 15 '12 at 6:55
+1 Nice. What software did you use for the diagrams? And ... where is Bezout? – Bill Dubuque Jun 15 '12 at 16:04
@BillDubuque You could check out GeoGebra for making images like this. – Honest Abe Aug 4 '12 at 3:47
@BillDubuque: I didn't see your message earlier. I used Intaglio, which, unfortunately, is only available on the Mac. – robjohn Aug 4 '12 at 4:41