Intersecting Circles IV

Construct a line tangential to two given circles.

We’re given two circles with centers $P$ and $Q$ and we’re asked to construct the points $S$ and $T$.

$\mbox{Given problem}$

We first connect $P$ and $Q$ and define $A$ to cut $\overline{PQ}$ into two segments of equal length. The normals $\overline{SP}$ and $\overline{TQ}$ we seek for are parallel to $\overline{RA}$ and therefore $\overline{RS}= \overline{RT}$

$\mbox{Connect}~P~ \mbox{and}~ Q~ \mbox{and construct}~ A$

Now we draw a circle around $A$ with radius $\overline{AP}= \overline{AQ}$.

$\mbox{We need to find}~B$

The intersection $B$ of the normal line through $R$ and $A$ and the just drawn circle satisfies $$ \overline{BS} = \overline{BT}~, $$ however, we do not have the normal yet. Before we proceed, we mention that if we translate the tangent along the normal, then $\overline{TH} = \overline{SP}$ and $$ \overline{PQ} \sin \alpha = \overline{HQ} $$ and $\overline{HQ}$ equals the difference of the radii of the two given circles.

$\mbox{Translate the normal}$

Now, we set $C$ to be the dropped perpendicular of $B$ onto the bottom line.

$C~ \mbox{is known and allows construction of} ~B$

From $$ \overline{AC} = \overline{AB} \sin \alpha = \overline{AQ} \sin \alpha = \frac{\overline{PQ}}{2} \, \sin \alpha $$ we get $$ \overline{AC} = \frac{\overline{HQ}}{2} $$ and that’s the crucial point, because $\overline{HQ}$ is known and that means that we can construct $C$, which is located in the middle of $D$ and $F$. The point $B$ is then just the intersection of the circle around $A$ and the vertical through $C$.

$\mbox{We show that}~ \overline{SB} = \overline{DB}$

As regards $\overline{DC}$, we read from the sketch that $$ \overline{PD} + 2 \overline{DC} + \overline{FQ} = \overline{PQ}~, $$ so $$ \overline{AQ} - \overline{DC} = \overline{AQ} - \frac{PQ}{2} + \frac{\overline{PD} +\overline{FQ}}{2} $$ and thus $$ \overline{AB} - \overline{DC} = \frac{\overline{PD} +\overline{FQ}}{2} = \frac{\overline{SP} +\overline{TQ}}{2} = \overline{RA} $$ From the next figure, where $\overline{GS} = \overline{DC}$, we see that $\bigtriangleup_{GBC}$ is isosceles and therefore $\bigtriangleup_{SBD}$ is as well.

$\mbox{Translation of tangent by}~\overline{DC}~\mbox{along the normal}$

We have finally constructed $S$ and $T$.

The same arguments apply to the case where $D$ moves from the right to left side of $P$, wich yields the inner tangent of the circles.

$\mbox{Inner and outer tangent}$

In the following animation, the two circles are shown, together with the tangents ($\overline{ST}$ above) and the three points $D$, $B$, and $F$. We see how the setup changes while $D$ is moving.
$\mbox{Euclidea 9.1 Demo}$