6.45*. Circles of radius $x$ and $y$ touch a circle of radius $R$, and the distance between the points of tangency is $a$. Calculate the length of the following common tangent to the first two circles:

a) external, if both tangencies are external or internal simultaneously;

b) internal, if one tangency is internal and the other is external.
6.45. Let both tangencies be external and $x \leqslant y$. A line passing through the center $O$ of the circle of radius $x$ parallel to the segment connecting the points of tangency intersects the circle of radius $y-x$ (with the center at the center of the circle of radius $y$) at points $A$ and $B$ (Fig. 6.9). Then $O A=a(R+x) / R$ and $O B=O A+a(y-x) / R=$ $=a(R+y) / R$. The square of the desired length of the common external tangent is

![](https://cdn.mathpix.com/cropped/2024_05_21_22b0ba5e8c5a1f49bd81g-159.jpg?height=241&width=246&top_left_y=259&top_left_x=1319)

Fig. 6.9

$$
O A \cdot O B=(a / R)^{2}(R+x)(R+y)
$$

Similar reasoning shows that if both tangencies are internal, then the square of the length of the external tangent is $(a / R)^{2}(R-x)(R-y)$, and if the circle of radius $x$ is externally tangent and the circle of radius $y$ is internally tangent, then the square of the length of the internal tangent is $(a / R)^{2}(R-y)(R+x)$.

3 Remark. In the case of internal tangency of the circles, it is assumed that $R>x$ and $R>y$.