1. Unit Six – Circles Review

2. Circle: Definition: A circle is the locus of points in a plane that are a fixed distance from a point called the center of the circle.  A has radius r = AB and diameter d = CD.

5. Cirmumference and Area – Make sure you know these formulas! C = πd or C = 2πr A = πr 2

6. Central Angle: A central angle is an angle whose vertex is the center of a circle. Sides are two radii of the circle. The sum of the measures of the central angles of a circle is 360°.

7. Arc An arc is an unbroken part of a circle created by the sides of a central angle. The measure of an arc is = to the measure of its corresponding central angle. Congruent Arcs have the same measure.

9. Adjacent arcs are arcs of the same circle that intersect at exactly one point. RS and ST are adjacent arcs.

10. Length of an Arc The length of an arc is a fraction of the circumference of the circle.

13. In a circle or congruent circles, two chords are congruent if and only if they are equidistant from the center of the circle

14. Find NP. Example: NP = 2(15) = 30 Hint - Use the Pythagorean Theorem to find NS first

15. Lesson Quiz: Part II 2. NGH 139 Find each measure. 3. HL 21

16. Lesson Quiz: Part III  12.9 4.  T   U, and AC = 47.2. Find PL to the nearest tenth.

17. An inscribed angle is an angle whose vertex is on a circle and whose sides contain chords of the circle. An intercepted arc consists of endpoints that lie on the sides of an inscribed angle and all the points of the circle between them. A chord or arc subtends an angle if its endpoints lie on the sides of the angle.

22. Lesson Quiz: Part I Find each measure. 1. RUS 2. a 25° 3

23. 3. Find the angle measures of ABCD. Lesson Quiz: Part II m A = 95° m B = 85° m C = 85° m D = 95°

25. Example 3: Problem Solving Application Early in its flight, the Apollo 11 spacecraft orbited Earth at an altitude of 120 miles. What was the distance from the spacecraft to Earth’s horizon rounded to the nearest mile? The answer will be the length of an imaginary segment from the spacecraft to Earth’s horizon. 1 Understand the Problem

26. 2 Make a Plan Draw a sketch. Let C be the center of Earth, E be the spacecraft, and H be a point on the horizon. You need to find the length of EH, which is tangent to  C at H. By Theorem 11-1-1, EH  CH. So ∆CHE is a right triangle.

27. Solve 3 EC = CD + ED = 4000 + 120 = 4120 mi EC 2 = EH² + CH 2 4120 2 = EH 2 + 4000 2 974,400 = EH 2 987 mi  EH Seg. Add. Post. Substitute 4000 for CD and 120 for ED. Pyth. Thm. Substitute the given values. Subtract 4000 2 from both sides. Take the square root of both sides.

29. Example 4: Using Properties of Tangents HK and HG are tangent to  F. Find HG. HK = HG 5a – 32 = 4 + 2a 3a – 32 = 4 2 segments tangent to  from same ext. point  segments . Substitute 5a – 32 for HK and 4 + 2a for HG. Subtract 2a from both sides. 3a = 36 a = 12 HG = 4 + 2(12) = 28 Add 32 to both sides. Divide both sides by 3. Substitute 12 for a. Simplify.

30. Lesson Quiz: Part III 3. Mount Mitchell peaks at 6,684 feet. What is the distance from this peak to the horizon, rounded to the nearest mile?  101 mi 4. FE and FG are tangent to  F. Find FG. 90