# Two Circles and Tangents from Their Centers Problem

Let $\Gamma_1$ and $\Gamma_2$ be two non overlapping circles with centers $O_1$ and $O_2$ respectively. From $O_1$, draw the two tangents to $\Gamma_2$ and let them intersect $\Gamma_1$ at points $A$ and $B$. Similarly, from $O_2$, draw the two tangents to $\Gamma_1$ and let them intersect $\Gamma_2$ at points $C$ and $D$.

Prove that $AB=CD$.

I've done some extensive angle chasing on this but have been unable to make any real progress. Can't decide whether $ABCD$ is supposed to be rectangle (as in my diagrams), a parallelogram or even a trapezium. There is a homothety taking one circle to the other but as far as I can see this doesn't help as we don't have a clearly defined center to this.

Any help/hints would be greatly appreciated.

• Have the radii the same length ? – Tony Piccolo Jul 30 '13 at 5:29
• Not necessarily, otherwise it would be trivial! – John Marty Jul 30 '13 at 5:32

## 1 Answer

$$\frac{|AM|}{|AP|} = \frac{|QT|}{|QP|} \qquad \implies \qquad |AB| = \frac{2\cdot \text{product of radii}}{\text{distance between centers}} = |CD|$$

Since the length of chord $AB$ is "symmetric" with respect to elements of $\bigcirc P$ and $\bigcirc Q$ ---that is, swapping the roles of the circles leaves the formula unchanged--- this length must match that of chord $CD$. • Thanks, the diagram is great. How do you create them? Is there an easy way? – John Marty Jul 30 '13 at 6:30
• @JohnMarty: I used a program called GeoGebra to create the diagram. – Blue Jul 30 '13 at 6:35
• I actually answered a duplicate of this question here. While I like the argument by "formula symmetry" above, the other answer may be helpful, as well. – Blue Dec 16 '15 at 17:30