# Tangent Line to Two Circles

Alignments to Content Standards: G-SRT.B.5

In the picture below, points $A$ and $B$ are the centers of two circles and they are collinear with point $C$. Also $D$ and $E$ lie on the two respective circles and they are also collinear with point $C$. Finally the two circles touch at a single point on $\overline{AB}$. What is $|BC|$? Explain.

## IM Commentary

The purpose of this task is to use similar triangles, setting up a proportion in order to calculate a side length. In order to establish the triangle similarity, students need to apply the result which says that if $Q$ is a point on a circle and $L$ is the tangent line to the circle at $Q$ then $L$ meets the radius of the circle at $Q$ perpendicularly. Once the appropriate proportion is selected, a further algebraic substitution is required in order to find the segment length indicated. The task provides a nice interplay between algebra and geometry, linked by similar triangles and the corresponding proportions to which they lead.

This task was adapted from problem #11 on the 2012 American Mathematics Competition (AMC) 10A Test. In the AMC exam question, the diagram was not given. For the 2012 AMC 10A, which was taken by 73703 students, the multiple choice answers for the problem had the following distribution:

 Choice Answer Percentage of Answers (A) 4 11.27 (B) 4.8 7.99 (C) 10.1 5.57 (D)* 12 17.22 (E) 14.4 2.53 Omit -- 55.40

Of the 73703 students: 36206, or 49%, were in 10th grade;25498 or 35%, were in 9th grade; and the remainder were below than 9th grade.

## Solution

Since $\overleftrightarrow{CD}$ is tangent to the circle with radius 5 at $D$, this means that $\angle ADC$ is a right angle. Similarly $\angle BEC$ is a right angle. We have $m(\angle BCE) = m(\angle ACD)$ because these are the same angles by the collinearity hypotheses given in the problem statement. By AA similarity, we have that $\triangle ADC$ is similar to $\triangle BEC$. This means that corresponding sides of these triangles are proportional so $$\frac{|BC|}{|AC|}=\frac{|BE|}{|AD|}.$$

We have been given that $|AD| = 5$ and $|BE| = 3$. We do not know $|AC|$ or $|BC|$ but we do know that $|AB| = 5+ 3 = 8$ because the two circles are tangent at the point on $\overline{AB}$ which is a distance of 5 from $A$ and 3 from $B$. We also know that $|AC| = |AB| + |BC| = 8 + |BC|$. Plugging all of this information into the above equation gives $$\frac{|BC|}{8 + |BC|} = \frac{3}{5}$$ Rewriting this equation gives $5|BC| = 24 + 3|BC|$ and so $|BC| = 12$.