# Common tangent to two circles with Ruler and Compass

Given two circles (centers are given) -- one is not contained within the other, the two do not intersect -- how to construct a line that is tangent to both of them? There are four such lines.

• Are you also given the centers of the circles? – Hagen von Eitzen Jan 12 '13 at 16:06

[I will assume you know how to do basic constructions, and not explain (say) how to draw perpendicular from a point to a line.]

If you're not given the center of the circles, draw 2 chords and take their perpendicular bisector to find the centers $O_1, O_2$.

Draw the line connecting the centers. Through each center, draw the perpendicular to $O_1O_2$, giving you the diameters of the circles that are parallel to each other. Connect up the endpoints of these diameters, and find the point of intersection. There are 2 ways to connect them up, 1 gives you the exterior center of expansion (homothety), the other gives you the interior center of expansion (homothety).

Each tangent must pass through one of these centers of expansion. From any center of expansion, draw two tangents to any circle. Extend this line, and it will be tangential to the other circle.

• What do you mean by 'Connect up the endpoints'? – Sigur Jan 12 '13 at 16:36
• Let the diameters that you get be $A_1 B_1$ and $A_2 B_2$ (where $A$'s are on different side of $O_1O_2$ than $B$'s. Connect up $A_1 A_2$ and $B_1B_2$, the intersection gives the exterior center, connect up $A_1 B_2$ and $A_2 B_1$, the intersection gives the interior center. – Calvin Lin Jan 12 '13 at 16:39
• @Calvin.Do you have a proof that each tangent must pass through one of centers of expansion – Adi Dani Jan 12 '13 at 17:24
• Depends on what you mean. Note that I didn't need to use the phrase 'center of expansion', but it provides the motivation behind what I'm doing. So, conversely, take a center of expansion of the two circles, draw $l$ that is tangential to $\Gamma_1$ at point $T_1$, passing through this point. Expand $\Gamma_1$, $T_1$, with a suitable ratio, that brings $\Gamma_1$ to $\Gamma_2$, and $T_1$ to $T_2$, which shows that $l$ is tangential to $\Gamma_2$ at $T_2$. ($l$ does not 'change' under the expansion, though the individual points do move further out). – Calvin Lin Jan 12 '13 at 17:30
• For alternative solutions see this link – Adi Dani Jan 12 '13 at 17:48