1. Circles – Modules 15

2. Module 15: Angles and Segments in Circles
Central Angle
Central Angle – The angle created by two
radii of the circle, with the vertex being the center of the circle. A
Measure of Arc AB = 105˚
105˚
The degree measure of a CENTRAL ANGLE and its INTERCEPTED ARC are the same! B

3. Module 15: Angles and Segments in Circles
Vocabulary - Chords
Chords – A segment whose endpoints lie on the outside of a circle.
Each of the segments in the diagram are chords.
Notice that even a diameter is a chord!
(A radius, however, is not a chord as it only intersects the outside of the circle at one point.)

4. Module 15: Angles and Segments in Circles
Inscribed Angle
Inscribed Angle – An angle intercepting an arc whose vertex is on the outside of the circle. (Notice that an inscribed angle is created by 2 chords!) A
105˚
IMPORTANT PROPERTY TO MEMORIZE:
The degree measure of an INSCRIBED ANGLE is ½ the measure of its intercepted arc. Note that this means an inscribed angle is also ½ the measure of its corresponding central angle!
105˚ B
Inscribed angle = ½ of 105˚ or 52.5˚
52.5˚

5. Module 15: Angles and Segments in Circles
Inscribed Angles
Inscribed Angles will always be ½ the measure of their intercepted arc, no matter where on the circle the vertex of the intercepted angle is! A a˚
If the measure of arc AB is a˚
Then ∠AXB, ∠AYB, and ∠AZB
each have a measure of ½ a˚ X B
½ a˚ Y Z

6. Module 15: Angles and Segments in Circles
Quadrilaterals Inscribed in Circles
Quadrilaterals Inscribed in Circles – These have a special relationship. Begin by examining two points on a circle, A and B.
100˚ A
If the measure of the minor Arc AB is 100˚ what is the measure of the major Arc AB?
360˚- 100˚ = 260˚
Notice that any two arcs created by the same two points on a circle must have measures that sum to 360˚! B
260˚

7. Module 15: Angles and Segments in Circles
Quadrilaterals Inscribed in Circles
What would the measure of each arc’s inscribed angles have to be?
100˚
If the measure of the minor Arc AB is 100˚ any inscribed angle that intercepts that arc must be ½ of the arc measure or 50˚
If the measure of the major Arc AB is 260˚ any inscribed angle that intercepts that arc must be ½ of the arc measure or 130˚
130˚ A B
50˚
260˚
130˚+ 50˚ = 180˚ → the angles are supplementary!

8. Module 15: Angles and Segments in Circles
Quadrilaterals Inscribed in Circles
Since the total angle measure of a quadrilateral must be 360˚, the remaining two angle (at point A and point B) must also be supplementary!
100˚
130˚ A
m∠A + m∠B = 180˚ B
50˚
260˚

9. Module 15: Angles and Segments in Circles
Quadrilaterals Inscribed in Circles
This can be stated generally as follows:
When a quadrilateral is inscribed in a circle, each pair of opposite angles MUST be supplementary. C A
m∠A + m∠B = 180˚
m∠C + m∠D = 180˚ B D

10. Module 15: Angles and Segments in Circles
Checkpoint – putting it all together: Inscribed Angles and Quadrilaterals.
60˚
Find the measure of all of the ?’s on the diagram. Don’t click ahead until you have them all! (There are multiple ways to solve for each measure). ?
80˚ ? ? ?
70˚ ?

11. Module 15: Angles and Segments in Circles
Checkpoint – putting it all together: Inscribed Angles and Quadrilaterals.
60˚
Find the measure of all of the ?’s on the diagram. Don’t click ahead until you have them all! (There are multiple ways to solve for each measure).
110˚
80˚
180˚- 80˚ = 100˚
2(100˚) - 60˚ = 140˚
180˚- 70˚ = 110˚
2(110˚) - 140˚ = 80˚
2(80˚) - 80˚ = 80˚
80˚
140˚
100˚
70˚
80˚

12. Module 15: Angles and Segments in Circles
A NOTE ABOUT DIAMETERS
Notice that a diameter cuts a circle in half, creating two 180˚ arcs. There will be times when this fact alone will be needed to solve a problem (you will not have enough angle measure information given without this knowledge.)
If one were to create an inscribe angle that intersected the circle at both ends of the diameter, the measure of that angle would be ½ of 180˚ or 90˚
Thus, an inscribed angle created with a diameter will always form a right triangle!
180˚
90˚
180˚

13. Module 15: Angles and Segments in Circles
Tangents
Tangent – A tangent line or segment is a line or segment that touches the circle at one and only one point, and never passes inside of a circle.
Point of tangency
Tangent
The point where a tangent intersects a circle is called the point of tangency.

14. Module 15: Angles and Segments in Circles
Tangents – Special relationships
Tangent and Radius – If a tangent line intersects with a radius (at the point of tangency), they will be perpendicular to each other, creating a right angle.
Point of tangency
Tangent
Radius
NOTE – This will be assumed knowledge on the EOC! There will never be a right angle labeled at this intersection, so you must memorize this fact.

15. Module 15: Angles and Segments in Circles
Tangents – Special relationships
The intersection of 2 tangents – If two lines are drawn tangent to a circle, they will connect and form a special kind of quadrilateral called a kite (unless the lines’ points of tangency are 180 degrees apart, in which case the lines will be parallel!)
I kite is a quadrilateral with two pairs of congruent, consecutive sides. On the next slide, the congruencies and other relationships of these “tangent kites” will be explained.

16. Module 15: Angles and Segments in Circles
Tangents – Special relationships
Opposite angles in “tangent” kites are SUPPLEMENTARY
Notice that one pair of opposite sides will ALWAYS be supplementary, as each angle created at the points of tangency are 90˚
Therefore, since the total angle measure of any quadrilateral is 360˚, the remaining two angles (labeled ‘a’ and ‘b’) must be supplementary as well. b a
m∠a + m∠b = 180˚

17. Module 15: Angles and Segments in Circles
Tangents – Special relationships
Connecting the center of the circle with the point where the two tangents intersect creates two congruent right triangles.
Since each triangle shares the hypotenuse (in green), each shares a radius, and each has a right angles, the triangles are congruent by the HL triangle congruence theorem.

18. Module 15: Angles and Segments in Circles
Tangents – Special relationships
By the concept that corresponding parts of congruent triangles are congruent (CPCTC), each angle and side of the triangles are congruent!
Notice that the green segment is in fact an angle bisector since it separates two pairs of congruent angles.

19. Module 15: Angles and Segments in Circles
Tangents – Solving Problems
Typical problem solving with tangents will involve a chord connecting the two points of tangency, thus creating an isosceles triangle (ABC) with CONGRUENT BASE ANGLES (∠ABC and ∠BAC).
In fact, there is another isosceles triangle (ADB) with congruent base angles (∠DAB and ∠DBA)! A C D B

20. Module 15: Angles and Segments in Circles
Tangents – Solving Problems
With one angle measure, it is now possible to solve for each angle in the diagram.
Try to find each angle measure if the measure of ∠BAC = 20˚
20˚ A
70˚
140˚ C D
40˚
20˚
70˚ B

21. Module 15: Angles and Segments in Circles
Intersecting Chords
When chords intersect inside of a circle, two very interesting relationships occur. We will not worry about these proofs here, simply memorize the relationships and practice problem solving. Watch the videos on the subject on the class website for information about the proofs….

22. Module 15: Angles and Segments in Circles
Intersecting Chords
Intersecting chords create four segments that have a proportional relationship. This relationships is called:
The Chord Chord Product Theorem:
If two chords intersect, the product of the created segments from each chord must be equal.
AQ ∙ QB = CQ ∙ QD C A Q D B

23. Module 15: Angles and Segments in Circles
Chord Chord Product Theorem Practice
Find the measure of the missing segment (labeled with an ‘x’) given the measures of the other three segments. C A
4 cm
6 cm
6 ∙ 3.4 = 4 ∙ x
20.4 = 4x
Divide both sides by 4….
x = 5.1 Q
3.4 cm
x cm D B

24. Module 15: Angles and Segments in Circles
Chord Chord Product Theorem Practice
Find the measure of the missing segment QB by solving for ‘x’. C A
8 cm
10 cm
4 ∙ 10 = 8 ∙ (x + 2)
40 = 8 ∙ (x + 2)
Divide both sides by 8….
5 = x + 2
Subtract 2 from both sides….
x = 3
Plug in 3 for x, and QR = 5 cm Q
4 cm
(x + 2) cm D B

25. Module 15: Angles and Segments in Circles
Intersecting Chords
Intersecting chords also create vertical angles whose measures are related to the chords they intercept:
The Intersecting Chords Angle Measurement Theorem:
If two chords intersect, then the measure of each angle formed is the average of the measures of the two intercepted arcs.
100˚+190˚ C A
100˚
145˚
145˚
190˚
= 145˚ D 2 B

26. Module 15: Angles and Segments in Circles
Secants - Vocabulary
Although none of the relationships between secants need to be explored in this course, it is useful to have a definition provided here:
A SECANT is a segment or line that intersects the circle at 2 points, and extends beyond the circle in one or more directions.
The math with chords inside of the circle is them same with the chord portions of secants.

27. Standard Form of the Equation of a Circle
The standard form for the equation of a circle is as follows:
(𝒙−𝒉) 𝟐 + (𝒚−𝒌) 𝟐 = 𝒓 𝟐
Where r represents the radius of the circle and the values of ‘h’ and ‘k’ represent the ‘x’ and ‘y’ values of the center of the circle (h,k). r
(h,k)

28. Standard Form of the Equation of a Circle
Example:
(𝒙−𝟑) 𝟐 + (𝒚+𝟐) 𝟐 = 25
Identify the radius, r = = 5
Identify the center, (3,-2)
Notice that the center of the circle has x and y values that are the OPPOSITE of the values in the original equation. Consider that if you plug the values of the center into the original equation, the x and y terms become zero. This is ALWAYS a good way to think about finding the center of the circle from the formula! 5
(3,-2)

29. Standard Form of the Equation of a Circle
Example – given the radius and the center, develop the equation:
What is the equation of a circle with:
Radius 9 and the Center at (-4,5)?
(𝒙+𝟒) 𝟐 + (𝒚−𝟓) 𝟐 = 81
Notice that 81 is just the radius of 9 squared
Notice again that the center of the circle has x and y values that are the OPPOSITE of the values in the original equation. 9
(-3,5)