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# 9.2: Tangent Lines

Difficulty Level: At Grade Created by: CK-12
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What if a line were drawn outside a circle that appeared to touch the circle at only one point? How could you determine if that line were actually a tangent?

### Tangent Lines

The tangent line and the radius drawn to the point of tangency have a unique relationship. Let’s investigate it here.

#### Investigation: Tangent Line and Radius Property

Tools needed: compass, ruler, pencil, paper, protractor

1. Using your compass, draw a circle. Locate the center and draw a radius. Label the radius AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*}, with A\begin{align*}A\end{align*} as the center.
2. Draw a tangent line, BC\begin{align*}\overleftrightarrow{BC}\end{align*}, where B\begin{align*}B\end{align*} is the point of tangency. To draw a tangent line, take your ruler and line it up with point B\begin{align*}B\end{align*}. Make sure that B\begin{align*}B\end{align*} is the only point on the circle that the line passes through.
3. Using your protractor, find mABC\begin{align*}m \angle ABC\end{align*}.

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.

To prove this theorem, the easiest way to do so is indirectly (proof by contradiction). Also, notice that this theorem uses the words “if and only if,” making it a biconditional statement. Therefore, the converse of this theorem is also true. Now let’s look at two tangent segments, drawn from the same external point. If we were to measure these two segments, we would find that they are equal.

Two Tangents Theorem: If two tangent segments are drawn from the same external point, then the segments are equal.

In A,CB¯¯¯¯¯¯¯¯\begin{align*}\bigodot A, \overline{CB}\end{align*} is tangent at point B\begin{align*}B\end{align*}. Find AC\begin{align*}AC\end{align*}. Reduce any radicals.

Solution: Because CB¯¯¯¯¯¯¯¯\begin{align*}\overline{CB}\end{align*} is tangent, AB¯¯¯¯¯¯¯¯CB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB} \bot \overline{CB}\end{align*}, making ABC\begin{align*}\triangle ABC\end{align*} a right triangle. We can use the Pythagorean Theorem to find AC\begin{align*}AC\end{align*}.

52+8225+6489AC=AC2=AC2=AC2=89\begin{align*}5^2+8^2 &= AC^2\\ 25+64 &= AC^2\\ 89 &= AC^2\\ AC &= \sqrt{89}\end{align*}

#### Calculating Length

Find DC\begin{align*}DC\end{align*}, in A\begin{align*}\bigodot A\end{align*}. Round your answer to the nearest hundredth.

DCDC=ACAD=8954.43\begin{align*}DC &= AC - AD\\ DC &= \sqrt{89}-5 \approx 4.43\end{align*}

#### Calculating Perimeter

Find the perimeter of ABC\begin{align*}\triangle ABC\end{align*}.

AE=AD,EB=BF\begin{align*}AE = AD, EB = BF\end{align*}, and CF=CD\begin{align*}CF = CD\end{align*}. Therefore, the perimeter of ABC=6+6+4+4+7+7=34\begin{align*}\triangle ABC=6+6+4+4+7+7=34\end{align*}.

We say that G\begin{align*}\bigodot G\end{align*} is inscribed in ABC\begin{align*}\triangle ABC\end{align*}. A circle is inscribed in a polygon, if every side of the polygon is tangent to the circle.

#### Solving for Unknown Values

Find the value of x\begin{align*}x\end{align*}.

Because AB¯¯¯¯¯¯¯¯AD¯¯¯¯¯¯¯¯\begin{align*}\overline{AB} \bot \overline{AD}\end{align*} and DC¯¯¯¯¯¯¯¯CB¯¯¯¯¯¯¯¯,AB¯¯¯¯¯¯¯¯\begin{align*}\overline{DC} \bot \overline{CB}, \overline{AB}\end{align*} and CB¯¯¯¯¯¯¯¯\begin{align*}\overline{CB}\end{align*} are tangent to the circle and also congruent. Set AB=CB\begin{align*}AB = CB\end{align*} and solve for x\begin{align*}x\end{align*}.

4x94xx=15=24=6\begin{align*}4x-9 &= 15\\ 4x &= 24\\ x &= 6\end{align*}

### Examples

#### Example 1

Determine if the triangle below is a right triangle. Explain why or why not.

To determine if the triangle is a right triangle, use the Pythagorean Theorem. 410\begin{align*}4 \sqrt{10}\end{align*} is the longest length, so we will set it equal to c\begin{align*}c\end{align*} in the formula.

82+10264+100 ? (410)2160\begin{align*}8^2+10^2 & \ ? \ \left( 4 \sqrt{10} \right)^2\\ 64+100 &\neq 160\end{align*}

ABC\begin{align*}\triangle ABC\end{align*} is not a right triangle. And, from the converse of the Tangent to a Circle Theorem, CB¯¯¯¯¯¯¯¯\begin{align*}\overline{CB}\end{align*} is not tangent to A\begin{align*}\bigodot A\end{align*}.

#### Example 2

Find the distance between the centers of the two circles. Reduce all radicals.

The distance between the two circles is AB\begin{align*}AB\end{align*}. They are not tangent, however, AD¯¯¯¯¯¯¯¯DC¯¯¯¯¯¯¯¯\begin{align*}\overline{AD} \bot \overline{DC}\end{align*} and DC¯¯¯¯¯¯¯¯CB¯¯¯¯¯¯¯¯\begin{align*}\overline{DC} \bot \overline{CB}\end{align*}. Let’s add BE¯¯¯¯¯¯¯¯\begin{align*}\overline{BE}\end{align*}, such that EDCB\begin{align*}EDCB\end{align*} is a rectangle. Then, use the Pythagorean Theorem to find AB\begin{align*}AB\end{align*}.

52+55225+30253050AC=AC2=AC2=AC2=3050=5122\begin{align*}5^2+55^2 &= AC^2\\ 25+3025 &= AC^2\\ 3050 &= AC^2\\ AC &= \sqrt{3050}=5\sqrt{122}\end{align*}

#### Example 3

If D\begin{align*}D\end{align*} and C\begin{align*}C\end{align*} are the centers and AE\begin{align*}AE\end{align*} is tangent to both circles, find DC\begin{align*}DC\end{align*}.

Because AE\begin{align*}AE\end{align*} is tangent to both circles, it is perpendicular to both radii and ABC\begin{align*}\triangle ABC\end{align*} and \begin{align*}\triangle DBE\end{align*} are similar. To find \begin{align*}DB\end{align*}, use the Pythagorean Theorem.

\begin{align*}10^2+24^2 &= DB^2\\ 100+576 &= 676\\ DB &= \sqrt{676}=26\end{align*}

To find \begin{align*}BC\end{align*}, use similar triangles.

\begin{align*}\frac{5}{10}=\frac{BC}{26} & \longrightarrow BC=13\\ DC=AB+BC &= 26+13=39\end{align*}

### Review

Determine whether the given segment is tangent to \begin{align*}\bigodot K\end{align*}.

Algebra Connection Find the value of the indicated length(s) in \begin{align*}\bigodot C\end{align*}. \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are points of tangency. Simplify all radicals.

1. \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are points of tangency for \begin{align*}\bigodot C\end{align*} and \begin{align*}\bigodot D\end{align*}, respectively.
1. Is \begin{align*}\triangle AEC \sim \triangle BED\end{align*}? Why?
2. Find \begin{align*}BC\end{align*}.
3. Find \begin{align*}AD\end{align*}.
4. Using the trigonometric ratios, find \begin{align*}m \angle C\end{align*}. Round to the nearest tenth of a degree.
2. Fill in the blanks in the proof of the Two Tangents Theorem. Given: \begin{align*}\overline{AB}\end{align*} and \begin{align*}\overline{CB}\end{align*} with points of tangency at \begin{align*}A\end{align*} and \begin{align*}C\end{align*}. \begin{align*}\overline{AD}\end{align*} and \begin{align*}\overline{DC}\end{align*} are radii. Prove: \begin{align*}\overline{AB} \cong \overline{CB}\end{align*}
Statement Reason
1.
2. \begin{align*}\overline{AD} \cong \overline{DC}\end{align*}
3. \begin{align*}\overline{DA} \bot \overline{AB}\end{align*} and \begin{align*}\overline{DC} \bot \overline{CB}\end{align*}
4. Definition of perpendicular lines
5. Connecting two existing points
6. \begin{align*}\triangle ADB\end{align*} and \begin{align*}\triangle DCB\end{align*} are right triangles
7. \begin{align*}\overline{DB} \cong \overline{DB}\end{align*}
8. \begin{align*}\triangle ABD \cong \triangle CBD\end{align*}
9. \begin{align*}\overline{AB} \cong \overline{CB}\end{align*}
1. From the above proof, we can also conclude (fill in the blanks):
1. \begin{align*}ABCD\end{align*} is a _____________ (type of quadrilateral).
2. The line that connects the ___________ and the external point \begin{align*}B\end{align*} _________ \begin{align*}\angle ADC\end{align*} and \begin{align*}\angle ABC\end{align*}.
2. Points \begin{align*}A, B, C\end{align*}, and \begin{align*}D\end{align*} are all points of tangency for the three tangent circles. Explain why \begin{align*}\overline{AT} \cong \overline{BT} \cong \overline{CT} \cong \overline{DT}\end{align*}.
3. Circles tangent at \begin{align*}T\end{align*} are centered at \begin{align*}M\end{align*} and \begin{align*}N\end{align*}. \begin{align*}\overline{ST}\end{align*} is tangent to both circles at \begin{align*}T\end{align*}. Find the radius of the smaller circle if \begin{align*}\overline{SN} \bot \overline{SM}\end{align*}, \begin{align*}SM=22, TN=25\end{align*} and \begin{align*}m \angle SNT=40^\circ\end{align*}.
4. Four circles are arranged inside an equilateral triangle as shown. If the triangle has sides equal to 16 cm, what is the radius of the bigger circle?
5. Circles centered at \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are tangent at \begin{align*}W\end{align*}. Explain why \begin{align*}A, B\end{align*} and \begin{align*}W\end{align*} are collinear.

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### Vocabulary Language: English Spanish

TermDefinition
diameter A chord that passes through the center of the circle. The length of a diameter is two times the length of a radius.
point of tangency The point where the tangent line touches the circle.
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.

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