7.2
Tangents to Circles
We now turn our attention to tangent lines and their relation to a circle. We defined a tangent to a circle in the last section as a line in the plane of the circle that intersects the circle in exactly one point. The fact that we have defined a tangent to a circle in this way does not mean that such a thing actually exists. We will start here by proving two things: first, that at each point of a circle there does indeed exist a line that is tangent to the circle; and second, that at each point of the circle there is just one such line.
Note that the definition of a tangent requires that the line be in the plane of the circle. Since virtually all of our work concerns the geometry of figures lying in a plane, we will from now on assume that whenever we have a line and a circle, the line lies in the plane of the circle; that is to say, from now on we will not mention the detail "in the plane of the circle."
Our first theorem shows that a tangent to a circle exists at each point of the circle.
- Theorem 7.5: If a line is drawn perpendicular to a radius of a circle at the endpoint of the radius, then that line is tangent to the circle.
Restatement: Let $P$ be a point of $\odot C$. If $\overleftrightarrow{PA}\perp\overline{PC}$, then $\overleftrightarrow{PA}$ is tangent to $\odot C$.
Prove: $\overleftrightarrow{PA}$ intersects $\odot C$ only at point $P$.
Proof (Indirect): There are two possibilities: Either $\overleftrightarrow{PA}$ intersects the circle only at $P$, or it intersects the circle a second time, at a point $X$ that is different from $P$. If that second possibility were to be the case, then $\triangle XPC$ would be a right triangle with hypotenuse $\overleftrightarrow{XC}$. Since the hypotenuse of a right triangle is longer than either of the legs, then $XC>PC$. That means that the distance from $X$ to the center of the circle would be greater than the radius of the circle, contradicting the assumption that $X$ lies on the circle. Thus the only point of $\overleftrightarrow{PA}$ that lies on the circle is the point $P$. That means that $\overleftrightarrow{PA}$ is a tangent to the circle.
This theorem actually gives us more than the mere existence of a tangent to a circle at each point of the circle. It tells us how to get that tangent — by constructing a perpendicular to the radius at the point on the circle. We now need to show that, in fact, this is the only way to get a tangent.
- Theorem 7.6: If a line is tangent to a circle at a point $P$ of the circle, then that line is perpendicular to the radius drawn to point $P$.
Given: $\overleftrightarrow{AP}$ is tangent to $\odot C$ at $P$.
Prove: $\overline{CP}\perp\overleftrightarrow{AP}$
Proof (Indirect): We are given that $\overleftrightarrow{AP}$is tangent to $\odot C$. Either $\overline{CP}\perp\overleftrightarrow{AP}$ or $\overline{CP}\not\perp\overleftrightarrow{AP}$. We will assume that $\overline{CP}\not\perp\overleftrightarrow{AP}$, and obtain a contradiction. If $\overline{CP}\not\perp\overleftrightarrow{AP}$, then there is another segment $\overline{CQ}$ that is perpendicular to $\overleftrightarrow{AP}$, with $Q$ on $\overleftrightarrow{AP}$. Let $R$ be the point on the ray opposite $\overrightarrow{QP}$ such that $\overline{QR}\cong\overline{QP}$.
Now $\triangle CQR\cong\triangle CQP$ by SAS, so $\overline{CR}\cong\overline{CP}$. Since $R$'s distance from $C$ is the radius of the circle, point $R$ lies on $\odot C$. But then $\overleftrightarrow{AP}$ has two points on the circle, and is thus not a tangent. This contradiction of the "given" for the problem means that the assumption $\overline{CP}\not\perp\overleftrightarrow{AP}$ is incorrect. Therefore $\overline{CP}\perp\overleftrightarrow{AP}$.
An important consequence of Theorem 7.6 is that at each point of a circle, there is just one tangent to the circle, because there is just one line in the plane of the circle that is perpendicular to the radius at the endpoint of the radius. Typically, when we want to do something with a line that is tangent to a circle at a point $T$, we put in the radius to the point of tangency, because, as we now know, that produces a right angle.
If $\overleftrightarrow{PT}$ is tangent to $\odot C$ at point $T$, then $\overline{PT}$ is called a tangent segment. The following is a very useful theorem with many applications.
- Theorem 7.7: Tangent segments to a circle from the same external point are congruent.
Restatement: If $\overline{PT}$ and $\overline{PQ}$ are both tangent segments drawn to $\odot C$ from point $P$, then $\overline{PT}\cong\overline{PQ}$. |
Proof:
\begin{array}{ l| l } Statement & Reasoning \\ \hline
1. \overline{PT} \text{ and } \overline{PQ}\text{ are tangent segments to }\odot C & 1. \text{ Given} \\
2. \overline{PT}\perp \overline{CT}\text{ and } \overline{PQ}\perp \overline{CQ}& 2. \text{ A tangent to a circle is perpendicular}\\
&\text{ to the radius drawn to the point of tangency.}\\
3. \angle PTC\text{ and } \angle PQC\text{ are right angles.} & 3. \text{ Def. of perpendicular}\\
4. \overline{CT}\cong \overline{CQ} & 4. \text{ All radii of a circle are congruent} \\
5. \overline{Cp}\cong \overline{CP} & 5. \text{ Reflexive}\\
6. \triangle CTP\cong\triangle CQP &6. \text{ H-L}\\
7. \overline{PT}\cong\overline{PQ} &7.\text{ CPCTC}
\end{array}
In Section 2.5, we proved that the bisector of an angle is the set of all points that are equidistant from the sides of the angle. This fact is intimately connected with the ideas that we have dealt with here. If a point $P$ is any point chosen on the bisector of $\angle ABC$, and if perpendiculars $\overline{PT}$ and $\overline{PQ}$ are drawn to the sides of the angle, those segments are congruent. Then a circle with its center at $P$ and radius $\overline{PT}$ will pass through $Q$. In addition, $\overrightarrow{BA}$ and $\overrightarrow{BC}$ will be tangents to $\odot P$.
This idea can be extended further. In Section 4.5, we proved that the bisectors of the three angles of a triangle are concurrent — they pass through a common point. This proof was based on the Angle Bisector Characterization Theorem. The point of concurrency has the special property that it is equidistant from all three sides of the triangle.
When we take that special point as the center of a circle and the common distance to the sides of the triangle as the radius, we obtain a circle that will touch each of the sides of the triangle. Since the distance from the point to each of the sides of the triangle is found by using a perpendicular to that side, each of the sides of the triangle is tangent to the circle. We say the circle is inscribed in the triangle and that the triangle is circumscribed about the circle. The point of concurrency of the angle bisectors of the angles of the triangle, being the center of the triangle's inscribed circle, is called the incenter of the triangle.
Example: What is the radius of the inscribed circle of a 3–4–5 triangle? |
\begin{align}
(3-r)+(4-r)&=5\\
7-2r&=5\\
2&=2r\\
r&=1
\end{align}
The radius of the inscribed circle is 1.
The approach used in the last example, that of using a variable to help organize a solution to a problem, can be useful in proofs as well as in numerical situations. Consider the next example.
Example: Pentagon $ABCDE$ is circumscribed about a circle. Prove that if $$AB+CD=BC+DE,$$ then the point of tangency of the fifth side $\overline{AE}$ is the midpoint of that side.
Proof:
Let the points of tangency be $P$, $Q$, $R$, $S$, and $T$, as
pictured. Assign lengths to the individual tangent segments as
shown. Using repeatedly the theorem that tangent segments from
the same external point are congruent, we know the lengths of the two
segments on $\overline{BC}$ and one of the segments on $\overline{DE}$,
so we fill them in. Let, $$BQ=x;\;QC=y;\;DS=z.$$ We are given that $AB+CD=BC+DE.$ This means that $$(w+x)+(y+z)=(x+y)+(z+t).$$ When we substract $x+y+z$ from both sides, we have $$w=t.$$ But $AT=AP=w$ and $TE=TS=t.$ Hence, $AT=ET$. Therefore, $T$ is the midpoint of $\overline{AE}$. |