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5. Tangent Line to a Circle

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Chapter 5

5.

Tangent Line to a Circle

Tangent lines to circles have been a topic of interest and study in geometry. This lesson delves into the methods of constructing a tangent line to a circle from an external point. It emphasizes the properties of tangents and how to ascertain that a drawn line is indeed a tangent. The lesson further elaborates on drawing a tangent line through an outer point using a compass and straightedge. The process involves connecting the external point to the circle's center, marking the midpoint, drawing a circle with the midpoint as the center, and finally drawing the tangents. The lesson also touches upon the real-life application of tangents, illustrating it with an example of a hammer throw in the Olympics.

Lesson Settings & Tools

	8 Theory slides
	11 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Tangent Line to a Circle

Slide of 8

Tangent lines have been defined and studied earlier in this course. In this lesson, methods for constructing a tangent line to a circle from an external point will be discussed.

Catch-Up and Review

Here is some recommended reading before getting started with this lesson.

Circles
Midpoint
Inscribed Angle Theorem
Right Angle
Inscribed Right Triangle Theorem
Tangent
Tangent to Circle Theorem

Challenge

Investigating Properties of Tangents to a Circle

Consider a circle with center O and point P outside of the circle. Using a straightedge and compass, can you construct a tangent to the circle through the given point?

Circle and External Point

Think about the properties of tangents. How can you justify that the line you draw is indeed the tangent?

Discussion

Drawing a Tangent Line to a Circle Through an Outer Point

Given a circle and a point outside the circle, a compass and straightedge can be used to draw a tangent line from the point to the circle.

There are four steps to construct a tangent line.

1

Draw PO

Using a straightedge, draw the segment connecting the point P and the center of the circle O.

2

Mark Midpoint M of PO

To locate the midpoint of PO, the perpendicular bisector of PO can be used. Recall that there are three steps to draw a perpendicular bisector.

The point where PO and its perpendicular bisector intersect is the midpoint of PO.

3

Draw a Circle with Center M and Radius MO

Next, place the compass tip on M and stretch the compass to O, or P. Then, draw the circle M.

4

Draw Tangents

The points where the circles intersect will be the points of tangency for the tangents drawn from P to the circle O.

The Inscribed Right Triangle Theorem can be used to justify why this construction works.

Why

Consider OA, a radius of ⊙ O.

In circle M, ∠ OAP is an inscribed angle on a diameter of ⊙ M. Since an inscribed angle opposite the diameter is a right angle, ∠ OAP is a right angle.

As can bee seen, OA ⊥ PA. The radius of ⊙ O is perpendicular to a line that passes through a point on the circle. Therefore, by the Tangent to Circle Theorem, PA is tangent to the circle.

Example

Using Tangents to a Circle to Construct an Eye

Kriz is learning a graphic program. By default, the program shows segment AB and circle O. The segment's endpoint A can be moved anywhere outside of ⊙ O. While endpoint B can be moved anywhere. An eye-like shape appears on the screen when AB is tangent to the circle. Give it a try!

Kriz cannot quite place point B in position to see the eye-like shape appear. Help Kriz out!

Answer

See solution.

Hint

How can a tangent line from a point outside of the given circle be constructed?

Solution

Since point A is a point outside ⊙ O, B should be the point of tangency in order for AB to be tangent to the circle. On the example shape, by extending AB, it can be observed that B is the point of tangency.

Constructing a tangent from an outer point will help locate the point of tangency for a tangent drawn from A. Recall the steps in constructing a tangent.

Connect the outer point and the center of the circle.
Find midpoint M of the segment drawn.
Draw a circle with center M and radius half the segment drawn.
Draw tangent.

In this case, point A is the outer point through which the tangent line is drawn. To get the example shape, move point A to the left as shown and then follow the steps.

As can be seen, the points where the circles intersect are the points of tangency. Therefore, point B should be on these points.

Discussion

Relationship Between Tangents and Lines of Symmetry

The lines of symmetry of a circle are the lines that passes through the center of the circle.

Suppose that a tangent line drawn from an outer point intersects a circle at A. When A is reflected across the line that passes through P and O, its image will also be a point of tangency for another tangent.

Discussion

External Tangent Congruence Theorem

Two tangent segments drawn from a common external point to the same circle are congruent.

For the above diagram, the following conditional statement holds true.

If AB and AC are tangent segments to ⊙ O, then AB ≅ AC.

Proof

Consider two triangles.

The triangle formed by the radius OB, the segment OA, and the tangent segment AB.
The triangle formed by the radius OC, the segment OA, and the tangent segment AC.

These two triangles can be visualized in the diagram.

Note that B and C are points of tangency. Therefore, by the Tangent to Circle Theorem, ∠ B and ∠ C are right angles. Consequently, △ ABO and △ ACO are right triangles.

Because all radii of the same circle are congruent, it can be said that OB and OC are congruent. Moreover, △ ABO and △ ACO share the same hypotenuse OA. By the Reflexive Property of Congruence, OA is congruent to itself.

Combining all of this information, it can be said that the hypotenuse and one leg of △ ABO are congruent to the hypotenuse and the corresponding leg of △ ACO.

OB ≅ OC OA ≅ OA

Therefore, by the Hypotenuse-Leg Theorem, △ ABO and △ ACO are congruent triangles. Since corresponding parts of congruent figures are congruent, it can be said that AB and AC are congruent.

AB ≅ AC

Example

External Tangent Congruence Theorem

In the diagram, all three segments are tangent to circle P.

The points D, E, and F are the points where the segments touch the circle. If AB= 12, BC=10, and CA=6, find AE.

Hint

Use the External Tangent Congruence Theorem.

Solution

From the graph, it can be seen that AD and AE are tangent segments with a common endpoint outside ⊙ P. By the External Tangent Congruence Theorem, AE and AD are congruent.

Smilarly, BE and BF, and CF and CD are congruent tangent segments. AD ≅ AE & & AD = AE [0.6em] BE ≅ BF & ⇔ & BE = BF [0.6em] CF ≅ CD & & CF = CD Using the Segment Addition Postulate and the given lengths, a system of equations with three equations and three unknowns can be written. AE+BE = AB BF+ CF = BC CD+ AD = CA ⇒ AE+BE = 12 & (I) BE+ CF = 10 & (II) CF+ AE = 6 & (III) To find AE, the Elimination Method can be used. Start by multiplying the second equation by - 1. - 1(BE+ CF) = - 1(10) ⇕ - BE- CF = - 10 Adding this equation to the first equation will eliminate BE.

AE+BE + ( - BE- CF )= 12 + ( - 10)

▼

Simplify

a+(- b)=a-b

AE+BE - BE - CF = 12 -10

Subtract term

AE - CF = 2

Since - CF and CF are additive inverses, adding this equation to the third equation will eliminate CF, and thus give AE.

CF+AE + ( AE- CF )= 6 + ( 2)

▼

Simplify

Remove parentheses

CF+AE + AE- CF = 6+2

Add and subtract fractions

2AE = 8

.LHS /2.=.RHS /2.

AE = 4

Closure

Tangents to Circles in Real Life

Imagine a superhero joining the Olympics to throw a hammer. An athlete would typically spin counterclockwise three or four (rarely five) times, then release the hammer. As viewed from above, the hammer travels on a path that is tangent to the circle created when the athlete spins. The diagram below shows the path of the superhero's hammer throw. See how the super hero fairs! Note, it is a not-to-scale drawing.

On August 30, 1986, in a packed stadium full of fans, Yuriy Sedykh set the world record with a throw of 86.74 meters. Perhaps a throw farther than the superhero!

Tangent Line to a Circle

Exercise 3.1

In △ ABC, we have AB=AC=12 and BC=8. Each segment is tangent to ⊙ P.

What is the radius of ⊙ P? Answer in exact form.

From the given information, we know that AB=AC=12 and BC=8. That means the triangle is isosceles. Additionally, we know all three sides are tangent to ⊙ P. Therefore, the radii PD, EP, and PF are perpendicular to the triangle's sides.

To determine the radius, we will analyze the area of the triangle.

Area △ ABC

The area of △ ABC is one half the product of its base and height. If we draw the height from the vertex angle A, it will bisect the base. Let's label the height h.

We can calculate h by using the Pythagorean Theorem.

The height of the triangle is 8sqrt(2) units. Using this value, we can evaluate the area of a triangle.

Area Inner Triangles

If we draw segments from the triangle's vertices to the center of the circle, we get three smaller inner triangles, each with a height of r.

Notice that the sum of the areas of these three triangles must be the same as the area of △ ABC. A_(△ ABC)=A_(△ BPC)+A_(△ BPA)+A_(△ APC) ⇓ 32sqrt(2)=1/2(8)r+1/2( 12)r+1/2( 12)r

Finding the Radius

Let's solve the equation to find the radius.

The radius of the circle P is 2sqrt(2) units.

Two semicircles, ⊙ O and ⊙ P, are inscribed in a square with a side of 10 centimeters. Both ⊙ O and ⊙ P are tangent to l.

What is the ratio of the smaller semicircle's area to the larger semicircle's area? Give the answer in exact form.

The area of a semicircle is half the area of a circle with the same radius. Typically, that is written as half the product of π and the radius squared. A=1/2π r^2

Area of the Greater Semicircle

The side of the square is 10 centimeters. The greater semicircle's radius is half the square's side. Therefore, r_1= 5 centimeters. We can substitute r into the formula to calculate its area. A=1/2π ( 5)^2 ⇒ A=1/2π (25)

Area of the Smaller Semicircle

According to the Tangent to Circle Theorem, l is perpendicular to both semicircles' radii at the point of tangency. This means OPB forms a right triangle.

The greater semicircle has a radius of 5 centimeters. If we call the radius of the smaller circle r_2, the triangle OPB will have legs of 5 and (10-r_2) centimeters and a hypotenuse of (5+r_2) centimeters. Let's add that information to the diagram.

To determine r_2, we will substitute a= 5, b= (10-r_2), and c= (5+r_2) into the Pythagorean Theorem. We will then solve for r_2.

Now that we know the value of r_2, we can determine the area of the smaller semicircle.

Ratio of Areas

Now that we know both semicircle's areas, we can determine the ratio of the smaller circle to the larger semicircle.

The ratio is 49.

Tangent Line to a Circle

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