Circles O, P, and Q are tangent as shown at points X, Y, If the larger gear has 30 teeth and the smaller gear has

CONSTRUCTIONS OF TANGENTS TO CIRCLES P a X b P X Z Y P c X Z Y W Figure 6.55 Construction 8 To construct a tangent to a circle at a point on the circle. PLAN: The strategy used in Construction 8 is based on Theorem 6.4.1. For Figure 6.55a, we will draw a radius extended beyond the circle. At the point on the circle point X in Figure 6.55, we construct the line perpendicular to The constructed line [ in Figure 6.55c] is tangent to circle P at point X. GIVEN: with point X on the circle [See Figure 6.55a.] CONSTRUCT: A tangent to at point X }P Í XW }P Í WX PX . CONSTRUCTION: Figure 6.55a: Consider and point X on . Figure 6.55b: Draw radius and extend it to form . Using X as the center and any radius length less than XP , draw two arcs to intersect at points Y and Z. Figure 6.55c: Complete the construction of the perpendicular line to at point X. From Y and Z, mark arcs with equal radii of length greater than XY. Calling the point of intersection W, draw , the desired tangent to at point X. }P Í XW PX PX PX PX }P }P EXAMPLE 1 Make a drawing so that points A, B, C, and D are on in that order. If tangents are constructed at points A, B, C, and D, what type of quadrilateral will be formed by the tangent segments if a and ? b all arcs , , , and are congruent? Solution a A rhombus all sides are congruent b A square all four are right ; all sides We now consider a more difficult construction. ⬵ ∠s ∠s DA ¬ CD ¬ BC ¬ AB ¬ mBC ¬ = mAD ¬ m AB ¬ = mCD ¬ }O 쮿 In the preceding construction, not shown is a radius of the smaller circle Q. In the larger circle M, is an inscribed angle that intercepts a semicircle. Thus, is a right angle and . Because the line drawn perpendicular to the radius of a circle at its endpoint on the circle is a tangent to the circle, is a tangent to circle Q. INEQUALITIES IN THE CIRCLE The remaining theorems in this section involve inequalities in the circle. ET ET ⬜ TQ ∠ETQ ∠ETQ QT M E Q b M E V Q T c Figure 6.56 E Q a Construction 9 To construct a tangent to a circle from an external point. GIVEN: and external point E [See Figure 6.56a.] CONSTRUCT: A tangent to , with T as the point of tangency CONSTRUCTION: Figure 6.56a: Consider and external point E. Figure 6.56b: Draw . Construct the perpendicular bisector of , to intersect at its midpoint M. Figure 6.56c: With M as center and MQ or ME as the length of radius, construct a circle. The points of intersection of circle M with circle Q are designated by T and V. Now draw , the desired tangent. NOTE: If drawn, would also be a tangent to . }Q EV ET EQ EQ EQ }Q }Q ET }Q Exs. 1–3 In a circle or in congruent circles containing two unequal central angles, the larger angle corresponds to the larger intercepted arc. THEOREM 6.4.2 GIVEN: with central angles and in Figure 6.57; PROVE: PROOF: In , . By the Central Angle Postulate, and By substitution, The converse of Theorem 6.4.2 follows, and it is also easily proved. m AB ¬ 7 mCD ¬ . m ∠2 = mCD ¬ . m ∠1 = mAB ¬ m ∠1 7 m∠2 }O m AB ¬ 7 mCD ¬ m ∠1 7 m∠2 ∠2 ∠1 }O D B A C O 1 2 Figure 6.57 In a circle or in congruent circles containing two unequal arcs, the larger arc corre- sponds to the larger central angle. THEOREM 6.4.3 GIVEN: In Figure 6.57, with and PROVE: The proof is left as Exercise 35 for the student. m ∠1 7 m∠2 m AB ¬ 7 mCD ¬ CD ¬ AB ¬ }O