CIRCLES Section 6 1 Vocabulary and Properties of

CIRCLES Section 6. 1 Vocabulary and Properties of Tangents.

Definitions O Circle: The set of points equidistant from one point called the center. O Radius: The distance from the center to any point on the circle. There are many radii in a circle. (½ of the diameter) O Diameter: The distance across a circle that goes through the center. (2 times the radius)

Labeled Drawing E A F B O A is the center O We call circles by the center, so this is circle A O Radius: AB O Diameter: EF O Name 2 other radii

Chords O Segments whose endpoints are on the circle. O DO NOT have to go through the center! O DE and BC are both chords. There are many in a circle! O A diameter IS a chord

Secant OA line that intersects a circle in two points OLine BC is a secant going through circle A.

Tangent O A line that intersects a circle in exactly one point. O BD is a line tangent to circle A. O Remember AB is a ____. O Point B is called the POINT of tangency.

Two types of Tangent Circles O Common Internal Tangent: A tangent line that intersects the segment that joins the centers of two circles. O Line EF is the common internal tangent to circle A and circle C.

Two types of Tangent Circles O Common External Tangent: Does not intersect the segment that joins the centers of two circles. O Line ED and line BF are common external tangents. Notice they DO NOT intersect segment AC.

Theorem 6. 1 O A line is tangent to a circle if and only if it is perpendicular to the radius drawn to the point of tangency. O Radius AB is perpendicular to tangent DC. O There are two right angles here. Name them.

Theorem 6. 2 O Tangent segments from a common external point are congruent. O Since line DE and line CE are both tangent to circle A, Segment DE is congruent to segment CE.

Check It Out! Example 4 b RS and RT are tangent to Q. Find RS. RS = RT 2 segments tangent to from same ext. point segments . Substitute n + 3 for RS n + 3 = 2 n – 1 and 2 n – 1 for RT. 4=n RS = 4 + 3 =7 Simplify. Substitute 4 for n. Simplify.

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