Circles formed by points of common tangency of two circles

Question

I was playing around in Geogebra with circles and their common tangents. For any two random circles (yellow), I noticed that the four points of tangency of direct common tangents form a circle, and so do the four points of tangency of the transverse common tangents.

The two circles (blue) are concentric, and their common centre M also happens to be the midpoint of the line segment joining the centres of the original two circles.

My question: how do I find the radii of the concentric circles, just using the radii of the original circles and the distance between them?

(Also, are there any more interesting properties of these circles, which have been documented somewhere?)

Answer 1

The well-known geometric construction of direct common tangent, which is shown in the diagram below, can be used to easily determine the dimeter of the mentioned circle that passes through the four points of tangency.

To construct the direct common tangents, we need to draw two auxiliary circles. One of those circles has the radius $\space R-r\space$ and the center $C_1$. The other is drawn with the radius $\dfrac{d}{2}$ and has the mod point of $C_1C_2$ as its center. One of their points of intersection is $Q$. Draw the segment $C_1Q$ and then extend it so that it cuts the circle $\Omega_1$ at $P_1$ as shown in the diagram. We also draw a line segment joining $Q$ and $C_2$. To get the direct common tangent, draw a line parallel to $QC_2$ through $P_1$ to through the circle $\Omega_2$ at $P_2$.

As per the construction, we can state that, $$C_1Q=R-r,\qquad\quad\tag{0}$$ $$P_1P_2 = QC_2 \qquad\text{and}\tag{1}$$ $$\measuredangle C_2QC_1 = 90^o.\qquad\space \tag{2}$$

Now, mark $N$ as the midpoint of $P_1P_2$. Since $M$ is the midpoint of $C_1C_2$, $MN$ is parallel to both $C_1P_1$ and $C_2P_2$. Therefore, we have, $$MN=\dfrac{1}{2}\left(C_1P_1+C_2P_2\right) =\dfrac{1}{2}\left(R+r\right).\tag{3}$$

Because $\measuredangle C_2QC_1 = 90^o$, the triangle $C_1C_2Q$ is right-angled. Using (0), we apply Pythagoras theorem to this triangle to obtain, $$QC_2^2=C_1C_2^2-C_1Q^2 = d^2-\left(R-r\right)^2.\tag{4}$$

Since $QC_2P_2P_1$ is a rectangle, using (4), we shall write, $$P_1P_2 = QC_2 = \sqrt{ d^2-\left(R-r\right)^2}.$$

$N$ is the midpoint of $P_1P_2$. Therefore, $$P_1N^2= \dfrac{1}{4}P_1P_2^2 = \dfrac{1}{4}\left( d^2-\left(R-r\right)\right)^2.\tag{5}$$

Finally, using (1) and (5), apply Pythagoras theorem to the right-angled triangle $MNP_1$ to express $MP_1$ in terms of the diameters of the two circles $\Omega_1$ and $\Omega_2$, and the distance $\space d\space$ between their centers. $$MP_1^2=MN^2+P_1N^2 = =\dfrac{1}{4}\left(R+r\right)^2 + \dfrac{1}{4}\left(d^2-\left(R-r\right)\right)^2$$ $$a=MP_1=\dfrac{1}{2}\sqrt{d^2+4Rr}$$

Added at OP's request - Transverse Common Tangent:

Answer 2

Let $C_1=(x_1,y_1)$ and $C_2=(x_2,y_2)$. Call the points of tangency on one of the direct tangents $A_1$ and $A_2$. Then because each of the segments $A_1C_1$ (of length $r_1$) and $A_2C_2$ (of length $r_2$) is perpendicular to that tangent, they are parallel. So if $O$ is the intersection of the two direct tangents, then $\triangle OA_1C_1$ and $\triangle OA_2C_2$ are similar right triangles. The former has one leg $r_1$ and hypotenuse $h$, And the latter has one leg $r_2$ and hypotenuse $$h+\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}.$$

Solving for $h$ will allow you to find $\angle C_1C_2A_2$, and thus the coordinates of $A_2$, which, of course, allows you to calculate the length of $MA_2$.

Answer 3

We use this property that the distances of points of intersection of the circle , passing the tangency points, with the line connecting the centers of two circles EN (in figure points P and O) from the centers of two circles are equal( in figure AP=CO=0.41). In particular case where the projection of point of tangency H on EN is coincided with reflect of point P about point A (the center of biger circle) we have AC, $AH=r_A$ and $CF=r_C$ as known of problem and we can write:

$\angle QCA=\angle P'HA$

$\Rightarrow \frac{AP=P'A}{AH=r_A}=\frac{r_A-r_C}{AC}$

in this way $AP=CO$ can be foud and the diameter of the circle passing the tangency points is:

$D=AC+2AP$

Update: to calculate the radius we can also use the power of E to the circle it's radius is required:

$EF\times EH=ES^2$

$tan HEA=tan QCA=\frac{r_A-r_C}{CQ}$

$sin QCA=\frac {r_A-r_C}{AC}$

These relations give measure of $QC=FH$

$\angle FEC=\angle QCA$

$EH=\frac{r_c}{tan FEC}$

$EH=EF+FH$

$ES^2=EF\times EH$

$EA=\frac{r_A}{tan HEA}$

$EC=\frac{r_c}{tan HEA}$

$ER=\frac{EA+EC}2$

Finally:

$r=RS=\sqrt{ER^2-ES^2}$