7.3
Tangents with Two Circles
In the last section, we dealt with tangents drawn to a single circle. When there are two circles, a variety of possibilities exist. Two different circles can intersect in two points, one point, or not at all. If they do not intersect, there could be one circle inside the other (perhaps concentric, perhaps not), or each outside the other.
Two circles that intersect in exactly one point are called tangent circles. However, we adopt a definition that is easier to work with.
- Definition: Two circles are tangent to each other (or simply, "are tangent") if and only if they are tangent to the same line at the same point.
In the figure below we have $\odot A$, $\odot C$, and $B$ on both circles. Move points $A$, $B$, and $C$. Try to create two internally tangent circles. Create two externally tangent circles. What is the relationship between the line of centers and points $A$, $B$, and $C$ in externally tangent circles? Internally tangent circles?
One particularly important fact about circles that are tangent to each other is that the point of tangency lies on the line of centers.
- Theorem 7.8:
If two circles are tangent to each other at a point $T$, then $T$ lies
on the line joining the centers of the circles.
Prove: $T$, $P$, and $Q$ are collinear.
Proof: Since $\odot P$ and $\odot Q$ are tangent to each other at point $T$, they have a common tangent line $\ell$ that passes through point $T$. Then $\overleftrightarrow{PT}\perp\ell$ and $\overleftrightarrow{QT}\perp\ell$. But there is only one line that is perpendicular to line $\ell$ at point $T$. Hence $\overrightarrow{PT}$ and $\overleftrightarrow{QT}$ must be the same line. Thus the three points $T$, $P$, and $Q$ are collinear.
Two circles can have a common tangent even if they are not tangent to each other. For instance, in the figures below, $\overleftrightarrow{PQ}$ is tangent to both $\odot A$ and $\odot B$.
There is an important difference between the two cases pictured. In the first case, the points of tangency lie on the same side of the line of centers; in the second case, the points of tangency lie on opposite sides of the line of centers. In the first case, $\overleftrightarrow{PQ}$ is a common external tangent to the two circles; in the second case, $\overleftrightarrow{PQ}$ is a common internal tangent. In both cases, the radii drawn to the points of tangency are perpendicular to the common tangent, and hence parallel to each other.
There are two typical problems having to do with common tangents. These are illustrated in the following examples.
Example: In the figure to the
right, $\overleftrightarrow{PQ}$ is tangent to the circles at $P$ and
$Q$. $AP=15$, $BQ=3$, and $AB=20$. How long is the common
tangent segment $\overline{PQ}$? |
Solution: We draw
the radii to the points of tangency, the line of centers, and the
segment from $B$ that is perpendicular to radius $\overline{AP}$.
This produces rectangle $PQBC$ and right $\triangle ABC$. The
hypotenuse $AB=20$ and the leg $AC=12$. With the Pythagorean
Theorem, we then compute $BC=16$. Since $PQBC$ is a rectangle,
$PQ=16$ as well. |
Example: In the figure to the right, $\overleftrightarrow{PQ}$ is tangent to the circles at $P$ and $Q$. $AP=15$, $BQ=3$, and $AB=20$. How long is the common tangent segment $\overline{PQ}$? |
Solution: Again we put in the
radii to the points of tangency. We could do the problem working
with similar triangles. However, it is easier to mimic the
solution of the last example, dropping a perpendicular from the center
of the smaller circle to the radius of the larger one. This gives
a large right $\triangle ABC$, with $AC=18$ (the sum of the radii of
the two circles). Then $$BC=\sqrt{20^2-18^2}=2\sqrt{19}.$$
Since $PQBC$ is a rectangle, $PQ=2\sqrt{19}$. |