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Geometry View details
5. Tangent Line to a Circle
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eCourses /
Geometry /
Chapter 5
5. 

Tangent Line to a Circle

Student Learning Objectives:
  • Apply the Tangent Congruence Theorem
  • Draw a tangent line to a circle through an outer point

Why
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.

Problem Solving Reasoning and Communication Error Analysis Modeling Using Tools Precision Pattern Recognition
Lesson Settings & Tools
8 Theory slides
11 Exercises - Grade E - A
Each lesson is meant to take 1-2 classroom sessions
Image Credits expand_more
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.

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

2
Mark Midpoint M of PO
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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
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Next, place the compass tip on M and stretch the compass to O, or P. Then, draw the circle M.

4
Draw Tangents
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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.

  1. Connect the outer point and the center of the circle.
  2. Find midpoint M of the segment drawn.
  3. Draw a circle with center M and radius half the segment drawn.
  4. 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
AddNeg

a+(- b)=a-b

AE+BE - BE - CF = 12 -10
SubTerm

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
RemovePar

Remove parentheses

CF+AE + AE- CF = 6+2
AddSubFrac

Add and subtract fractions

2AE = 8
DivEqn

.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!



Level 1
Level 2
Level 3
Tangent Line to a Circle
Exercise 1.1
Is BC tangent to the circle?
triangle ABC and circle A
triangle ABC and a circle A
triangle ABC with its small leg being the diameter of circle A

Suppose a segment is perpendicular to a radius at its endpoint on a circle's circumference. In that case, we know by the Tangent to Circle Theorem that this segment is tangent to the circle. Therefore, to determine if BC is tangent to the circle, we can use the Pythagorean Theorem to evaluate if △ ABC is a right triangle.

Since we obtained a true statement, the triangle is, in fact, a right triangle. This means BC is tangent to the circle ⊙ A.


Notice that we only know part of the length of AC. However, the unknown part is also the circle's radius, which means AC must be the sum of the known part and the circle's radius.

Now we can use the Pythagorean Theorem to determine if △ ABC is a right triangle.

As we can see, △ ABC is a right triangle. Therefore, BC is tangent to the circle.


As in previous parts, we will use the Pythagorean Theorem to evaluate if BC is tangent to the circle. Notice that one of the triangle's legs is the diameter of the circle. Since we have been given the radius, we find the diameter by multiplying this value by 2.

Let's substitute the legs and hypotenuse in the Pythagorean Theorem and determine if it makes a true statement or not.

Since the Pythagorean Theorem is not satisfied, △ ABC is not a right triangle. Therefore, BC can not be tangent to the circle.

E
Hint & Answer
S
Solution
BC is tangent to ⊙ A. Find the radius of ⊙ A. Answer in exact form.
A circle with a right triangle ABC where B is on the circles circumference
A circle and triangle ABC where B is tangent to the circle

Notice that △ ABC is a right triangle. This triangle has legs of r and 24 units and a hypotenuse of (r+18) units. Let's mark these lengths in the diagram.

If we substitute the hypotenuse and the legs into the Pythagorean Theorem, we can solve for r.

We found that the radius r is 7.

As in Part A, we have a right triangle. Let's identify its legs and hypotenuse.

As in Part A, we will substitute the length of the triangle's legs and hypotenuse into the Pythagorean Theorem to determine the radius r.

The radius r is 163.

E
Hint & Answer
S
Solution
Determine the value of x if we know that AB and AD are tangent segments to the circle.
circle with two tangents to the circle which coincides at the common exterior point of A
circle with tangents to the circle which intersect at A

According to the External Tangent Congruence Theorem, if two line segments drawn from the same exterior point are tangent to a circle, then they are congruent.

With this information, we can write an equation that we can solve for x.


As in Part A, we will equate the expressions for the two tangents and solve for x.


E
Hint & Answer
S
Solution

What is the perimeter of TVUX if ⊙ Y is inscribed in the quadrilateral.

Circle Y inscribed in the quadrilateral TVWX

According to the External Tangent Congruence Theorem, tangents drawn from a common external point are congruent segments. This means XD and XA are congruent, as are UD and UC. Let's add this information to the diagram.

Now we can calculate XD, and thereby also XA, by subtracting UD from XU.

Let's add the length of XD and XA to the diagram.

Now we can determine the perimeter of TVUX by adding all of the sides.

E
Hint & Answer
S
Solution
The following circle represents a cross section of Earth where h is an arbitrary height above the ground and d is the line of sight across the horizon. The radius of the Earth is about 6400 km.

person standing h km above the Earth watching d km across the horizon
 What is d if h=5 kilometers? Round to the nearest integer.

What is h if d=500 kilometers? Round to one decimal place.

We know that the radius of Earth is about 6400 kilometers. We want to know the distance d one can see over the horizon when we are 5 kilometers above the Earth.

We see that ET= 6400 and EH= 6400+ 5. To find d, we will use the Pythagorean Theorem.

The distance a person can see when standing 5 km above the Earth is about 253 km.


Now that we know the distance one sees over the horizon. Let's update our illustration.

By substituting the legs and hypotenuse into the Pythagorean Theorem, we can determine h.

To solve this equation, we will use the Quadratic Formula.

When we see 500 kilometers across the horizon, we are about 19.5 kilometers above the ground!

E
Hint & Answer
S
Solution

Tangent Line to a Circle

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