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# 9.1: Parts of Circles & Tangent Lines

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Define the parts of a circle.
• Discover the properties of tangent lines.

## Review Queue

1. Find the equation of the line with m=25\begin{align*}m = \frac{2}{5}\end{align*} and y\begin{align*}y-\end{align*}intercept of 4.
2. Find the equation of the line with m=2\begin{align*}m = -2\end{align*} and passes through (4, -5).
3. Find the equation of the line that passes though (6, 2) and (-3, -1).
4. Find the equation of the line perpendicular to the line in #2 and passes through (-8, 11).

Know What? The clock to the right is an ancient astronomical clock in Prague. It has a large background circle that tells the local time and the “ancient time” and the smaller circle rotates to show the current astrological sign. The yellow point is the center of the larger clock. How does the orange line relate to the small and large circle? How does the hand with the moon on it relate to both circles?

## Defining Terms

Circle: The set of all points that are the same distance away from a specific point, called the center.

The center of the circle is point A\begin{align*}A\end{align*}. We call this circle, “circle A\begin{align*}A\end{align*},” and it is labeled A\begin{align*}\bigodot A\end{align*}.

Radii (the plural of radius) are line segments. There are infinitely many radii in any circle and they are all equal.

Radius: The distance from the center to the circle.

Chord: A line segment whose endpoints are on a circle.

Diameter: A chord that passes through the center of the circle.

Secant: A line that intersects a circle in two points.

The tangent ray TP\begin{align*}\overrightarrow{TP}\end{align*} and tangent segment TP¯¯¯¯¯¯¯\begin{align*}\overline{TP}\end{align*} are also called tangents.

The length of a diameter is two times the length of a radius.

Tangent: A line that intersects a circle in exactly one point.

Point of Tangency: The point where the tangent line touches the circle.

Example 1: Find the parts of A\begin{align*}\bigodot A\end{align*} that best fit each description.

b) A chord

c) A tangent line

d) A point of tangency

e) A diameter

f) A secant

Solution:

a) HA¯¯¯¯¯¯¯¯\begin{align*}\overline{HA}\end{align*} or AF¯¯¯¯¯¯¯¯\begin{align*}\overline{AF}\end{align*}

b) CD¯¯¯¯¯¯¯¯, HF¯¯¯¯¯¯¯¯\begin{align*}\overline{CD}, \ \overline{HF}\end{align*}, or DG¯¯¯¯¯¯¯¯\begin{align*}\overline {DG}\end{align*}

c) BJ\begin{align*}\overleftrightarrow{BJ}\end{align*}

d) Point H\begin{align*}H\end{align*}

e) HF¯¯¯¯¯¯¯¯\begin{align*}\overline{HF}\end{align*}

f) BD\begin{align*}\overleftrightarrow{BD}\end{align*}

## Coplanar Circles

Example 2: Draw an example of how two circles can intersect with no, one and two points of intersection. You will make three separate drawings.

Solution:

Tangent Circles: When two circles intersect at one point.

Concentric Circles: When two circles have the same center, but different radii.

Congruent Circles: Two circles with the same radius, but different centers.

If two circles have different radii, they are similar. All circles are similar.

Example 3: Determine if any of the following circles are congruent.

Solution: From each center, count the units to the circle. It is easiest to count vertically or horizontally. Doing this, we have:

Radius of ARadius of BRadius of C=3 units=4 units=3 units\begin{align*}\text{Radius of} \ \bigodot A & = 3 \ units\\ \text{Radius of} \ \bigodot B & = 4 \ units\\ \text{Radius of} \ \bigodot C & = 3 \ units\end{align*}

From these measurements, we see that AC\begin{align*}\bigodot A \cong \bigodot C\end{align*}.

Notice the circles are congruent. The lengths of the radii are equal.

## Internally & Externally Tangent

If two circles are tangent to each other, then they are internally or externally tangent.

Internally Tangent Circles: When two circles are tangent and one is inside the other.

Externally Tangent Circles: When two circles are tangent and next to each other.

Internally Tangent

Externally Tangent

If circles are not tangent, they can still share a tangent line, called a common tangent.

Common Internal Tangent: A line that is tangent to two circles and passes between the circles.

Common External Tangent: A line that is tangent to two circles and stays on the top or bottom of both circles.

Common Internal Tangent

Common External Tangent

Let’s investigate a tangent line and the radius drawn to the point of tangency.

Investigation 9-1: 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*}. B\begin{align*}B\end{align*} must be the only point on the circle that the line passes through.
3. 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.

BC\begin{align*}\overleftrightarrow{BC}\end{align*} is tangent at point B\begin{align*}B\end{align*} if and only if BCAB¯¯¯¯¯¯¯¯\begin{align*}\overleftrightarrow{BC} \perp \overline{AB}\end{align*}.

This theorem uses the words “if and only if,” making it a biconditional statement, which means the converse of this theorem is also true.

Example 4: 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: CB¯¯¯¯¯¯¯¯\begin{align*}\overline{CB}\end{align*} is tangent, so AB¯¯¯¯¯¯¯¯CB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB} \perp \overline{CB}\end{align*} and ABC\begin{align*}\triangle ABC\end{align*} a right triangle. 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*}

Example 5: Find DC\begin{align*}DC\end{align*}, in A\begin{align*}\bigodot A\end{align*}. Round your answer to the nearest hundredth.

Solution: DC=ACADDC=8954.43\begin{align*}DC = AC - AD\!\\ DC = \sqrt{89}-5 \approx 4.43\end{align*}

Example 6: Determine if the triangle below is a right triangle.

Solution: Again, use the Pythagorean Theorem. 410\begin{align*}4\sqrt{10}\end{align*} is the longest side, so it will be c\begin{align*}c\end{align*}.

82+102 64+100? (410)2160\begin{align*}8^2+10^2 \ & ? \ \left ( 4\sqrt{10} \right )^2\\ 64+100 & \ne 160\end{align*}

ABC\begin{align*}\triangle ABC\end{align*} is not a right triangle. From this, we also find that CB¯¯¯¯¯¯¯¯\begin{align*}\overline{CB}\end{align*} is not tangent to A\begin{align*}\bigodot A\end{align*}.

Example 7: Find AB\begin{align*}AB\end{align*} in A\begin{align*}\bigodot A\end{align*} and B\begin{align*}\bigodot B\end{align*}. Reduce the radical.

Solution: AD¯¯¯¯¯¯¯¯DC¯¯¯¯¯¯¯¯\begin{align*}\overline{AD} \perp \overline{DC}\end{align*} and DC¯¯¯¯¯¯¯¯CB¯¯¯¯¯¯¯¯\begin{align*}\overline{DC} \perp \overline{CB}\end{align*}. Draw in BE¯¯¯¯¯¯¯¯\begin{align*}\overline{BE}\end{align*}, so EDCB\begin{align*}EDCB\end{align*} is a rectangle. 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*}

## Tangent Segments

Theorem 9-2: If two tangent segments are drawn from the same external point, then they are equal.

BC¯¯¯¯¯¯¯¯\begin{align*}\overline{BC}\end{align*} and DC¯¯¯¯¯¯¯¯\begin{align*}\overline{DC}\end{align*} have C\begin{align*}C\end{align*} as an endpoint and are tangent; BC¯¯¯¯¯¯¯¯DC¯¯¯¯¯¯¯¯\begin{align*}\overline{BC} \cong \overline{DC}\end{align*}.

Example 8: Find the perimeter of ABC\begin{align*}\triangle ABC\end{align*}.

Solution: 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*}.

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.

Example 9: If D\begin{align*}D\end{align*} and A\begin{align*}A\end{align*} are the centers and AE\begin{align*}AE\end{align*} is tangent to both circles, find DC\begin{align*}DC\end{align*}.

Solution: AE¯¯¯¯¯¯¯¯DE¯¯¯¯¯¯¯¯\begin{align*}\overline{AE} \perp \overline{DE}\end{align*} and AE¯¯¯¯¯¯¯¯AC¯¯¯¯¯¯¯¯\begin{align*}\overline{AE} \perp \overline{AC}\end{align*} and ABCDBE\begin{align*}\triangle ABC \sim \triangle DBE\end{align*}.

To find DB\begin{align*}DB\end{align*}, use the Pythagorean Theorem.

102+242100+576DB=DB2=676=676=26\begin{align*}10^2+24^2&=DB^2\\ 100+576&=676\\ DB&=\sqrt{676}=26\end{align*}

To find BC\begin{align*}BC\end{align*}, use similar triangles. 510=BC26BC=13. DC=AB+BC=26+13=39\begin{align*}\frac{5}{10}=\frac{BC}{26} \longrightarrow BC=13. \ DC=AB+BC=26+13=39\end{align*}

Example 10: Algebra Connection Find the value of x\begin{align*}x\end{align*}.

Solution: AB¯¯¯¯¯¯¯¯CB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB} \cong \overline{CB}\end{align*} by Theorem 9-2. 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*}

Know What? Revisited The orange line is a diameter of the smaller circle. Since this line passes through the center of the larger circle (yellow point), it is part of one of its diameters. The “moon” hand is a diameter of the larger circle, but a secant of the smaller circle.

## Review Questions

• Questions 1-9 are similar to Example 1.
• Questions 10-12 are similar to Example 2.
• Questions 13-17 are similar to Example 3.
• Questions 18-20 are similar to Example 6.
• Questions 21-26 are similar to Example 4, 5, 7, and 10.
• Questions 27-31 are similar to Example 9.
• Questions 32-37 are similar to Example 8.
• Question 38 and 39 use the proof of Theorem 9-2.
• Question 40 uses Theorem 9-2.

Determine which term best describes each of the following parts of P\begin{align*}\bigodot P\end{align*}.

1. KG¯¯¯¯¯¯¯¯\begin{align*}\overline{KG}\end{align*}
2. FH\begin{align*}\overleftrightarrow{FH}\end{align*}
3. KH¯¯¯¯¯¯¯¯¯\begin{align*}\overline{KH}\end{align*}
4. E\begin{align*}E\end{align*}
5. BK\begin{align*}\overleftrightarrow{BK}\end{align*}
6. CF\begin{align*}\overleftrightarrow{CF}\end{align*}
7. A\begin{align*}A\end{align*}
8. JG¯¯¯¯¯¯¯\begin{align*}\overline{JG}\end{align*}
9. What is the longest chord in any circle?

Copy each pair of circles. Draw in all common tangents.

Coordinate Geometry Use the graph below to answer the following questions.

1. Find the radius of each circle.
2. Are any circles congruent? How do you know?
3. Find all the common tangents for B\begin{align*}\bigodot B\end{align*} and C\begin{align*}\bigodot C\end{align*}.
4. C\begin{align*}\bigodot C\end{align*} and E\begin{align*}\bigodot E\end{align*} are externally tangent. What is CE\begin{align*}CE\end{align*}?
5. Find the equation of CE¯¯¯¯¯¯¯¯\begin{align*}\overline{CE}\end{align*}.

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

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

A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*} are points of tangency for C\begin{align*}\bigodot C\end{align*} and D\begin{align*}\bigodot D\end{align*}.

1. Is AECBED\begin{align*}\triangle AEC \sim \triangle BED\end{align*}? Why?
2. Find CE\begin{align*}CE\end{align*}.
3. Find BE\begin{align*}BE\end{align*}.
4. Find ED\begin{align*}ED\end{align*}.
5. Find BC\begin{align*}BC\end{align*} and AD\begin{align*}AD\end{align*}.

A\begin{align*}\bigodot A\end{align*} is inscribed in BDFH\begin{align*}BDFH\end{align*}.

1. Find the perimeter of BDFH\begin{align*}BDFH\end{align*}.
2. What type of quadrilateral is BDFH\begin{align*}BDFH\end{align*}? How do you know?
3. Draw a circle inscribed in a square. If the radius of the circle is 5, what is the perimeter of the square?
4. Can a circle be inscribed in a rectangle? If so, draw it. If not, explain.
5. Draw a triangle with two sides tangent to a circle, but the third side is not.
6. Can a circle be inscribed in an obtuse triangle? If so, draw it. If not, explain.
7. Fill in the blanks in the proof of Theorem 9-2. Given: AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*} and CB¯¯¯¯¯¯¯¯\begin{align*}\overline{CB}\end{align*} with points of tangency at A\begin{align*}A\end{align*} and C\begin{align*}C\end{align*}. AD¯¯¯¯¯¯¯¯\begin{align*}\overline{AD}\end{align*} and DC¯¯¯¯¯¯¯¯\begin{align*}\overline{DC}\end{align*} are radii. Prove: AB¯¯¯¯¯¯¯¯CB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB} \cong \overline{CB}\end{align*}
Statement Reason
1.
2. AD¯¯¯¯¯¯¯¯DC¯¯¯¯¯¯¯¯\begin{align*}\overline{AD} \cong \overline{DC}\end{align*}
3. DA¯¯¯¯¯¯¯¯AB¯¯¯¯¯¯¯¯\begin{align*}\overline{DA} \perp \overline{AB}\end{align*} and DC¯¯¯¯¯¯¯¯CB¯¯¯¯¯¯¯¯\begin{align*}\overline{DC} \perp \overline{CB}\end{align*}
4. Definition of perpendicular lines
5. Connecting two existing points
6. ADB\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. Fill in the blanks, using the proof from #38.
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 ABC\end{align*}.
2. Points \begin{align*}A, \ B,\end{align*} and \begin{align*}C\end{align*} are points of tangency for the three tangent circles. Explain why \begin{align*}\overline{AT} \cong \overline{BT} \cong \overline{CT}\end{align*}.

1. \begin{align*}y = \frac{2}{5} x + 4\end{align*}
2. \begin{align*}y = -2x + 3\end{align*}
3. \begin{align*}{\;} \ m = \frac{2 - (-1)}{6 - (-3)} = \frac{3}{9} = \frac{1}{3}\!\\ {\;} \ y = \frac{1}{3} x+b \rightarrow \text{plug in} (6, 2)\!\\ {\;} \ 2 = \frac{1}{3} (6)+b\!\\ {\;} \ 2 =2+b \rightarrow b=0\!\\ {\;} \ y = \frac{1}{3} x\end{align*}
4. \begin{align*}{\;} \ \ m_\perp = -3\!\\ {\;} \quad 11 = -3(-8)+b\!\\ {\;} \quad 11 = 24+b \rightarrow b=-13\!\\ {\;} \quad \ \ y = -3x-13 \end{align*}

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