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# 9.4: Angles of Secants and Tangents

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Find the relationship between a radius and a tangent to a circle.
• Find the relationship between two tangents drawn from the same point.
• Find the measures of angles formed by secants and tangents.

## Tangent to a Circle

A tangent to a circle is a line that intersects the circle at exactly one point. This intersection point is called the point of tangency.

• A ____________________________ line intersects a circle exactly once.
• A point of __________________________________ is a point where a tangent line intersects a circle.

Tangent to a Circle Theorem

A tangent line is always at a right angle to the radius of the circle at the point of tangency.

• A tangent line and a __________________________ meet at a right angle.

Proof

We will prove the theorem on the previous page by contradiction (which starts by assuming the opposite of what you are trying to prove).

Start by making a drawing. AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*} is a radius of the circle. A\begin{align*}A\end{align*} is the center of the circle and B\begin{align*}B\end{align*} is the point of intersection between the radius and the tangent line. Another name for point B\begin{align*}B\end{align*} is the point of tangency.

Assume that the tangent line is not perpendicular to the radius.

There must be another point C\begin{align*}C\end{align*} on the tangent line such that AC¯¯¯¯¯¯¯¯\begin{align*}\overline{AC}\end{align*} is perpendicular to the tangent line. Therefore, in the right triangle ΔACB,AB¯¯¯¯¯¯¯¯\begin{align*}\Delta ACB, \overline{AB}\end{align*} is the hypotenuse and AC¯¯¯¯¯¯¯¯\begin{align*}\overline{AC}\end{align*} is a leg of the triangle.

However, this is not possible because AC>AB\begin{align*}AC > AB\end{align*}.

(Note that AC=\begin{align*}AC =\end{align*} the length of the radius +DC\begin{align*}+ DC\end{align*})

Since our assumption led us to a contradiction, this means that our assumption was incorrect. Therefore, the tangent line must be perpendicular to the radius of the circle.

Since the tangent to a circle and the radius of the circle make a right angle with each other, we can often use the Pythagorean Theorem to find the length of missing segments.

Complete the sentence:

This proof tells us that every tangent line is perpendicular to

________________________________________________________________________

_______________________________________________________________________.

Example 1

In the figure, CB¯¯¯¯¯¯¯¯\begin{align*}\overline{CB}\end{align*} is tangent to the circle. Find the length of CD¯¯¯¯¯¯¯¯\begin{align*}\overline{CD}\end{align*}.

Since CB¯¯¯¯¯¯¯¯\begin{align*}\overline{CB}\end{align*} is tangent to the circle, then CB¯¯¯¯¯¯¯¯  AB¯¯¯¯¯¯¯¯\begin{align*}\overline{CB} \ \bot \ \overline{AB}\end{align*}.

This means that ΔABC\begin{align*}\Delta ABC\end{align*} is a right triangle and we can apply the Pythagorean Theorem to find the length of AC¯¯¯¯¯¯¯¯\begin{align*}\overline{AC}\end{align*}. In this triangle, AC¯¯¯¯¯¯¯¯\begin{align*}\overline{AC}\end{align*} is the hypotenuse:

(AC)2(AC)2AC=(AB)2+(BC)2=52+82=25+64=89=899.43 in\begin{align*}(AC)^2 &= (AB)^2 + (BC)^2\\ (AC)^2 &= 5^2 + 8^2 = 25 + 64 = 89\\ AC &= \sqrt{89} \approx 9.43 \ in\end{align*}

AD¯¯¯¯¯¯¯¯\begin{align*}\overline{AD}\end{align*} is congruent to AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*} because they are each a radius of the circle!

CD¯¯¯¯¯¯¯¯\begin{align*}\overline{CD}\end{align*} is part of AC¯¯¯¯¯¯¯¯\begin{align*}\overline{AC}\end{align*}: CD=ACAD9.4354.43 in\begin{align*}CD = AC - AD \approx 9.43 - 5 \approx 4.43 \ in\end{align*}.

1. True or flase: A tangent line touches a circle at exactly one point, and it is perpendicular to the radius at that point.

2. Draw a picture that shows whether the statement in #1 above is true or false.

\begin{align*}{\;\;}\end{align*}

\begin{align*}{\;\;}\end{align*}

\begin{align*}{\;\;}\end{align*}

\begin{align*}{\;\;}\end{align*}

3. What is the name of the point where the tangent line intersects the radius?

\begin{align*}{\;\;}\end{align*}

\begin{align*}{\;\;}\end{align*}

\begin{align*}{\;\;}\end{align*}

\begin{align*}{\;\;}\end{align*}

Example 2

Mark is standing at the top of Mt. Whitney, which is 14,500 feet tall. The radius of the Earth is approximately 3,960 miles. (There are 5,280 feet in one mile.) How far can Mark see to the horizon?

We start by drawing the figure above. Without a drawing, this problem can be tricky! With a drawing, you can see that this is a simple tangent and radius problem.

The distance to the horizon is given by the line segment CB¯¯¯¯¯¯¯¯\begin{align*}\overline{CB}\end{align*}.

First, we must convert the height of the mountain from feet into miles:

14500 feet1 mile5280 feet=2.75 miles\begin{align*}14500 \ feet \cdot \frac{1 \ mile}{5280 \ feet} = 2.75 \ miles\end{align*}

Since CB¯¯¯¯¯¯¯¯\begin{align*}\overline{CB}\end{align*} is tangent to the Earth, ΔABC\begin{align*}\Delta ABC\end{align*} is a right triangle and we can use the Pythagorean Theorem. What are the lengths of each side of the triangle?

AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*} is the radius and a leg of the triangle, and AB=\begin{align*}AB = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}

AC¯¯¯¯¯¯¯¯\begin{align*}\overline{AC}\end{align*} is the hypotenuse of the triangle.

AC=radius+height of Mt. Whitney=+\begin{align*}AC = \text{radius} + \text{height of Mt. Whitney} = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} + \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}

CB¯¯¯¯¯¯¯¯\begin{align*}\overline{CB}\end{align*} is the tangent, which is the missing side of the triangle.

(CB)2+(AB)2(CB)2+(3960)2(CB)2CB=(AC)2=(3960+2.75)2=(3960+2.75)2(3960)2=21787.56=21787.56147.6 miles\begin{align*}(CB)^2 + (AB)^2 & = (AC)^2\\ (CB)^2 + (3960)^2 & = (3960 + 2.75)^2\\ (CB)^2 & = (3960 + 2.75)^2 - (3960)^2 = 21787.56\\ CB & = \sqrt{21787.56} \approx 147.6 \ miles\end{align*}

Mark can see about 147.6 miles to the horizon.

Converse of a Tangent to a Circle Theorem

If a line is perpendicular to the radius of a circle at its outer endpoint, then the line is tangent to the circle.

Remember, a converse is an if-then statement where the hypothesis (part after “if”) and the conclusion (part after “then”) are switched.

Proof

We will prove this theorem by contradiction.

Since the line is perpendicular to the radius at its outer endpoint it must touch the circle at point B\begin{align*}B\end{align*}. For this line to be tangent to the circle, it must only touch the circle at this point and no other.

Assume that the line also intersects the circle at point C\begin{align*}C\end{align*}.

Since both AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*} and AC¯¯¯¯¯¯¯¯\begin{align*}\overline{AC}\end{align*} are radii of the circle, AB¯¯¯¯¯¯¯¯AC¯¯¯¯¯¯¯¯\begin{align*}\overline{AB} \cong \overline{AC}\end{align*}, and ΔABC\begin{align*}\Delta ABC\end{align*} is isosceles.

This means that mABC=mACB=90\begin{align*}m \angle ABC = m \angle ACB = 90^\circ\end{align*}.

However, it is impossible to have two right angles in the same triangle. We arrived at a contradiction so our assumption must be incorrect. We conclude that line BC¯¯¯¯¯¯¯¯\begin{align*}\overline{BC}\end{align*} is tangent to the circle at point \begin{align*}B\end{align*}.

Example 3

Determine whether \begin{align*}\overline{LM}\end{align*} is tangent to the circle:

\begin{align*}\overline{LM}\end{align*} is tangent to the circle if \begin{align*}\overline{LM} \ \bot \ \overline{MN}\end{align*}.

To show that \begin{align*}\Delta LMN\end{align*} is a right triangle we use the Converse of the Pythagorean Theorem, which means that we show the Pythagorean Theorem works with the given numbers:

\begin{align*}&(LM)^2 + (MN)^2 = (LN)^2\\ &\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}^2 + \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}^2 = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}^2\\ &64 + 36 = 100 \ \text{yes, this is correct!}\end{align*}

The lengths of the sides of the triangle satisfy the Pythagorean Theorem, so \begin{align*}\overline{LM}\end{align*} is perpendicular to \begin{align*}\overline{MN}\end{align*} and is therefore tangent to the circle.

Measure of Tangent-Chord Angle Theorem

The measure of an angle formed by a chord and a tangent that intersect on the circle is half the measure of the intercepted arc.

• When a chord intersects a tangent on a circle, the angle formed is ____________ the measure of the intercepted arc.

In other words:

\begin{align*}m \angle FAB = \frac{1}{2} m \widehat{ACB}\end{align*} and \begin{align*}m \angle EAB = \frac{1}{2} m \widehat{ADB}\end{align*}

Proof

Draw the radii of the circle to points \begin{align*}A\end{align*} and \begin{align*}B\end{align*}:

\begin{align*}\Delta AOB\end{align*} is isosceles because \begin{align*}\overline{BO}\end{align*} and \begin{align*}\overline{OA}\end{align*} are both radii, so:

\begin{align*}m \angle BAO = m \angle ABO & = \frac{1}{2}(180^\circ - m \angle AOB)\\ & = 90^\circ - \frac{1}{2} m \angle AOB\end{align*}

We also know that \begin{align*}m \angle BAO + m \angle EAB = 90^\circ\end{align*} because \begin{align*}\overline{FE}\end{align*} is tangent to the circle.

\begin{align*}(90^\circ - \frac{1}{2} m \angle AOB )+ m \angle EAB = 90^\circ\end{align*}

Which means that \begin{align*}m \angle EAB = \frac{1}{2} m \angle AOB\end{align*}

Since \begin{align*}\angle AOB\end{align*} is a central angle that corresponds to \begin{align*}\widehat{ADB}\end{align*} then, \begin{align*}m \angle EAB = \frac{1}{2} m \widehat{ADB}\end{align*}.

This completes the proof.

In your own words, describe the angle that is formed when a chord intersects a tangent at a point on a circle.

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

Example 4

Find the values of \begin{align*}a, b\end{align*} and \begin{align*}c\end{align*}:

First we find angle \begin{align*}a\end{align*}:

\begin{align*}50^\circ + 45^\circ + m \angle a & = 180^\circ\\ 95^\circ + m \angle a & = 180^\circ\\ m \angle a & = 85^\circ\end{align*}

Next we find angles \begin{align*}b\end{align*} and \begin{align*}c\end{align*}.

Using the Measure of the Tangent-Chord Theorem (which says that the intercepted arc is double the angle formed by the __________________ and the ___________________ ), we conclude that:

\begin{align*}m \widehat{AB} = 2(45^\circ) = 90^\circ\end{align*} and \begin{align*}m \widehat{AC} = 2(50^\circ) = 100^\circ\end{align*}

Therefore,

\begin{align*}m \angle b & = \frac{1}{2} (m \widehat{AC}) = \frac{1}{2}(100^\circ) = 50^\circ \quad \text{and}\\ m \angle c & = \frac{1}{2} (m \widehat{AB}) = \frac{1}{2}(90^\circ) = 45^\circ\end{align*}

We can confirm this because all three angles in a triangle add up to \begin{align*}180^\circ\end{align*}:

\begin{align*}&m \angle a + m \angle b + m \angle c = 180^\circ\\ &\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} + \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} + \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} = 180^\circ\end{align*}

And yes, \begin{align*}180^\circ = 180^\circ\end{align*}!

Angles Outside a Circle Theorem

The measure of an angle formed by two secants drawn from a point outside the circle is equal to half the difference of the measures of the intercepted arcs.

In other words, \begin{align*}m \angle a = \frac{1}{2} (y^\circ - x^\circ)\end{align*}

• To find the measure of \begin{align*}\angle a\end{align*} in the diagram above, first take the arc with a measure of __________ and subtract the arc with a measure of ___________. Then take half of that difference.

This theorem also applies for an angle formed by two tangents to the circle drawn from a point outside the circle and for an angle formed by a tangent and a secant drawn from a point outside the circle:

The diagram above is an example of the angle formed by: a _______________________ and a _______________________

The diagram above is an example of the angle formed by: a _______________________ and a _______________________

Proof

Draw a line to connect points \begin{align*}A\end{align*} and \begin{align*}B\end{align*}:

Statements Reasons
1. \begin{align*}m \angle DBA = \frac{1}{2} x^\circ\end{align*} 1. Inscribed angle
2. \begin{align*}m \angle BAC = \frac{1}{2} y^\circ\end{align*} 2. Inscribed angle
3. \begin{align*}m \angle BAC = m \angle DBA + m \angle a\end{align*} 3. The measure of an exterior angle in a triangle is equal to the sum of the measures of the remote interior angles.
4. \begin{align*}\frac{1}{2} y^\circ = \frac{1}{2} x^\circ + m \angle a\end{align*} 4. Substitution
5. \begin{align*}m \angle a = \frac{1}{2} (y^\circ - x^\circ)\end{align*} 5. Subtract, simplify, and factor.

Example 5

Find the measure of angle \begin{align*}x\end{align*}:

\begin{align*}m \angle x = \frac{1}{2}(220^\circ - 54^\circ) = 83^\circ\end{align*}

1. When two secants, two tangents, or a tangent and a secant intersect outside of a circle, the angle formed makes the same relationship with the intercepted arcs. What is this relationship? Describe.

\begin{align*}{\;\;}\end{align*}

\begin{align*}{\;\;}\end{align*}

\begin{align*}{\;\;}\end{align*}

\begin{align*}{\;\;}\end{align*}

2. Draw a picture of two secants that intersect outside of a circle:

\begin{align*}{\;\;}\end{align*}

\begin{align*}{\;\;}\end{align*}

\begin{align*}{\;\;}\end{align*}

\begin{align*}{\;\;}\end{align*}

3. Draw a picture of two tangents that intersect outside of a circle:

\begin{align*}{\;\;}\end{align*}

\begin{align*}{\;\;}\end{align*}

\begin{align*}{\;\;}\end{align*}

\begin{align*}{\;\;}\end{align*}

4. Draw a picture of a secant and a tangent that intersect outside of a circle:

\begin{align*}{\;\;}\end{align*}

\begin{align*}{\;\;}\end{align*}

\begin{align*}{\;\;}\end{align*}

\begin{align*}{\;\;}\end{align*}

5. True or false: The measure of an angle formed by two secants (or two tangents or a secant and a tangent) outside a circle is the same as the average of the intercepted arcs.

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