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 $S_3$.
 A: Drop perpendiculars to $A$ and $B$ as in the diagram below and let $F$ and $G$ be the respective intersections with $S_1$ and $S_2$.
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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.
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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.
