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Tangent Lines

Lines perpendicular to the radius drawn to the point of tangency.

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Tangent Lines to Circles

\begin{align*}\overleftrightarrow{DC}\end{align*} and \begin{align*}\overleftrightarrow{CE}\end{align*} are tangent to circle \begin{align*}A\end{align*} at points \begin{align*}D\end{align*} and \begin{align*}E\end{align*} respectively. What type of quadrilateral is \begin{align*}ADCE\end{align*}? Can you find \begin{align*}m\angle DCE\end{align*}?

Tangent Lines to Circles

When a line intersects a circle in exactly one point the line is said to be tangent to the circle or a tangent of the circle. Below, line \begin{align*}l\end{align*} is tangent to the circle at point \begin{align*}P\end{align*}.

You will prove that if a tangent line intersects a circle at point \begin{align*}P\end{align*}, then the tangent line is perpendicular to the radius drawn to point \begin{align*}P\end{align*}.

From any point outside a circle, you can drawn two lines tangent to the circle. You will learn how to construct these lines in problems later. Below, from point \begin{align*}C\end{align*} both lines \begin{align*}l\end{align*} and \begin{align*}m\end{align*} are tangent to circle \begin{align*}A\end{align*}.

In the second problem, you will show that in this situation, \begin{align*}\overline{PC}\cong \overline{CQ}\end{align*}. In third problem, you will show that \begin{align*}\angle PAQ\end{align*} and \begin{align*}\angle PCQ\end{align*} are supplementary.

Let's look at a few example problems.  

1. Line \begin{align*}l\end{align*} is tangent to circle \begin{align*}A\end{align*} at point \begin{align*}P\end{align*}. Prove that line \begin{align*}l\end{align*} is perpendicular to \begin{align*}\overline{AP}\end{align*}.

This proof relies on the fact that the shortest distance from a point to a line is along the segment perpendicular to the line.

Consider a point \begin{align*}Q\end{align*} on line \begin{align*}l\end{align*} but not on circle \begin{align*}A\end{align*}. \begin{align*}AQ>AP\end{align*}, because \begin{align*}Q\end{align*} is outside circle \begin{align*}A\end{align*}. This means that the shortest distance from line \begin{align*}l\end{align*} to point \begin{align*}A\end{align*} is from point \begin{align*}P\end{align*} to point \begin{align*}A\end{align*}. Therefore, \begin{align*}\overline{AP}\end{align*} must be perpendicular to line \begin{align*}l\end{align*}.

2. From point \begin{align*}C\end{align*}, both lines \begin{align*}l\end{align*} and \begin{align*}m\end{align*} are tangent to circle \begin{align*}A\end{align*}. Show that \begin{align*}\overline{PC}\cong \overline{QC}\end{align*}. What does this mean in general?

Draw a segment connecting \begin{align*}A\end{align*} and \begin{align*}C\end{align*}. Note that \begin{align*}\angle AQC\end{align*} is also a right angle.

\begin{align*}\overline{AC}\cong \overline{AC}\end{align*} by the reflexive property and \begin{align*}\overline{PA}\cong \overline{QA}\end{align*} because they are both radii of the circle. This means that \begin{align*}\Delta APC\cong \Delta AQC\end{align*} by \begin{align*}HL\cong\end{align*}. \begin{align*}\overline{PC}\cong \overline{QC}\end{align*} because the segments are corresponding parts of congruent triangles.

\begin{align*}\overline{PC}\end{align*} and \begin{align*}\overline{QC}\end{align*} are known as tangent segments. In general, two tangent segments to a circle from the same point outside the circle will always be congruent.

3. From point \begin{align*}C\end{align*}, both lines \begin{align*}l\end{align*} and \begin{align*}m\end{align*} are tangent to circle \begin{align*}A\end{align*}. Show that \begin{align*}\angle PAQ\end{align*} and \begin{align*}\angle PCQ\end{align*} are supplementary. What does this mean in general?

\begin{align*}\angle ACQ\end{align*}  is a right angle because line \begin{align*}m\end{align*} is tangent to circle \begin{align*}A\end{align*} at point \begin{align*}Q\end{align*}. The sum of the measures of the interior angles of a quadrilateral is \begin{align*}360^\circ\end{align*}. This means that \begin{align*}m\angle PAQ+m\angle PCQ=360^\circ-90^\circ-90^\circ=180^\circ\end{align*}. Therefore, \begin{align*}\angle PAQ\end{align*} and \begin{align*}\angle PCQ\end{align*} are supplementary.

In general, the angle between two lines tangent to a circle from the same point will be supplementary to the central angle created by the two tangent lines.

Examples

Example 1

Earlier, you were given a problem about tangent lines to a circle. 

\begin{align*}\overleftrightarrow{DC}\end{align*} and \begin{align*}\overleftrightarrow{CE}\end{align*} are tangent to circle \begin{align*}A\end{align*} at points \begin{align*}D\end{align*} and \begin{align*}E\end{align*} respectively. What type of quadrilateral is \begin{align*}ADCE\end{align*}? Can you find \begin{align*}m\angle DCE\end{align*}?

\begin{align*}\overline{DA}\end{align*} and \begin{align*}\overline{EA}\end{align*} are both radii of the circle, so they are congruent. \begin{align*}\overline{DC}\end{align*} and \begin{align*}\overline{EC}\end{align*} are both tangent segments to the circle from the same point \begin{align*}(C)\end{align*}, so they are congruent. The quadrilateral has two pairs of adjacent congruent segments so it is a kite.

\begin{align*}m \widehat{DE}=360^\circ-238^\circ=122^\circ\end{align*}. The means \begin{align*}m\angle DAE=122^\circ\end{align*}. Because \begin{align*}\overleftrightarrow{DC}\end{align*} and \begin{align*}\overleftrightarrow{CE}\end{align*} are tangent to circle \begin{align*}A\end{align*}, you know that \begin{align*}\angle DAE\end{align*} and \begin{align*}\angle DCE\end{align*} are supplementary. \begin{align*}m\angle DCE=180^\circ-122^\circ=58^\circ\end{align*}.

In the following questions, you will learn how to construct lines tangent to a circle from a given point.

Example 2

Use your compass and straightedge (or another construction device) to construct a circle and a point not on the circle. Label the center of the circle \begin{align*}A\end{align*} and the point not on the circle \begin{align*}C\end{align*}.

Source: CK-12 Foundation. Author: Kaitlyn Spong and Laura Guerin
License: CC BY-NC 3.0

Example 3

Find the midpoint of \begin{align*}\overline{AC}\end{align*} and label it \begin{align*}M\end{align*}. Construct a circle centered at \begin{align*}M\end{align*} that passes through both \begin{align*}A\end{align*} and \begin{align*}C\end{align*}.

Construct the perpendicular bisector of \begin{align*}\overline{AC}\end{align*} in order to find its midpoint.

Source: CK-12 Foundation. Author: Kaitlyn Spong and Laura Guerin
License: CC BY-NC 3.0

Then construct a circle centered at point \begin{align*}M\end{align*} that passes through point \begin{align*}C\end{align*}. The circle should also pass through point \begin{align*}A\end{align*}.

Source: CK-12 Foundation. Author: Kaitlyn Spong and Laura Guerin
License: CC BY-NC 3.0

Example 4

Find the points of intersection of circle \begin{align*}M\end{align*} and circle \begin{align*}A\end{align*}. Label the points of intersection \begin{align*}P\end{align*} and \begin{align*}Q\end{align*}. Connect point \begin{align*}C\end{align*} with point \begin{align*}P\end{align*} and point \begin{align*}C\end{align*} with point \begin{align*}Q\end{align*}. Why are \begin{align*}\overleftrightarrow{CP}\end{align*} and \begin{align*}\overleftrightarrow{CQ}\end{align*} tangent lines?

Find the points of intersection and connect them with point \begin{align*}C\end{align*}.

Note that \begin{align*}\overline{AC}\end{align*} is a diameter of circle \begin{align*}M\end{align*}, so it divides circle \begin{align*}M\end{align*} into two semicircles. \begin{align*}\angle APC\end{align*} and \begin{align*}\angle AQC\end{align*} are inscribed angles of these semicircles, so they must be right angles. \begin{align*}\overline{PC}\end{align*} meets radius \begin{align*}\overline{AP}\end{align*} at a right angle, so \begin{align*}\overline{PC}\end{align*} is tangent to circle \begin{align*}A\end{align*}. Similarly, \begin{align*}\overline{QC}\end{align*} meets radius \begin{align*}\overline{AQ}\end{align*} at a right angle, so \begin{align*}\overline{QC}\end{align*} is tangent to circle \begin{align*}A\end{align*}.

Review

1. What is a tangent line?

For all pictures below, assume that lines that appear tangent are tangent.

Use the image below for #2-#3.

2. Draw in \begin{align*}\overline{AP}\end{align*} and find its length.

3. Find \begin{align*}AC\end{align*}.

Use the image below for #4-#7.

4. Find \begin{align*}m\angle CAQ\end{align*}.

5. Find \begin{align*}QC\end{align*}.

6. Find \begin{align*}AQ\end{align*}.

7. Find \begin{align*}PC\end{align*}.

Use the image below for #8-#9.

8. Find \begin{align*}m\widehat{PQ}\end{align*}.

9. Find \begin{align*}m\widehat{PEQ}\end{align*}.

Use the image below for #10-#11. 62% of the circle is purple.

10. Find the measure of the purple arc.

11. Find the measure of angle \begin{align*}\theta\end{align*}.

Use the image below for #12-#13.

12. Make a conjecture about how \begin{align*}\Delta ABI\end{align*} and \begin{align*}\Delta HGI\end{align*} are related.

13. Prove your conjecture from #12.

14. Use construction tools of your choice to construct a circle and a point not on the circle. Then, construct two lines tangent to the circle that pass through the point. Hint: Look at the Guided Practice questions for the steps for this construction.

15. Justify why your construction from #14 created tangent lines.

Review (Answers)

To see the Review answers, open this PDF file and look for section 8.7. 

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Vocabulary

Tangent

The tangent of an angle in a right triangle is a value found by dividing the length of the side opposite the given angle by the length of the side adjacent to the given angle.

Tangent to a Circle Theorem

A line is tangent to a circle if and only if the line is perpendicular to the radius drawn to the point of tangency.

Two Tangent Theorem

The Two-Tangent Theorem states that if two tangent segments are drawn to one circle from the same external point, then they are congruent.

Image Attributions

  1. [1]^ Source: CK-12 Foundation. Author: Kaitlyn Spong and Laura Guerin; License: CC BY-NC 3.0
  2. [2]^ Source: CK-12 Foundation. Author: Kaitlyn Spong and Laura Guerin; License: CC BY-NC 3.0
  3. [3]^ Source: CK-12 Foundation. Author: Kaitlyn Spong and Laura Guerin; License: CC BY-NC 3.0

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