Two circles with radii a and b touch each other externally such that $\theta$ is the angle between the direct common tangents $(a > b \geq 2)$, then
- $\theta = 2 \cos^{-1} \displaystyle \left ( \frac{a-b}{a+b}\right )$
- $\theta = 2\tan^{-1} \displaystyle \left ( \frac{a+b}{a-b}\right )$
- $\theta = 2\sin^{-1} \displaystyle \left ( \frac{a+b}{a-b}\right )$
- $\theta = 2 \sin^{-1} \displaystyle \left ( \frac{a-b}{a+b}\right )$
For two circles of radii a and b touching externally, the angle theta between the direct common tangents is given by the formula sin(theta/2) = (a-b)/(a+b). Thus, theta = 2 * arcsin((a-b)/(a+b)).
Draw a line connecting the centers of the two externally touching circles with radii a and b, and drop perpendiculars from both centers to the direct common tangent. This creates a right triangle where the hypotenuse is the sum of the radii (a plus b) and the leg opposite half the angle theta is the difference of the radii (a minus b). Using trigonometric ratios, the sine of half theta equals the opposite side over the hypotenuse, giving sin(theta divided by 2) equals (a minus b) divided by (a plus b). Multiplying by 2 gives the final angle relation, theta equals 2 times the inverse sine of (a minus b) divided by (a plus b).