# Circles in Neutral Geometry

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1 Everything we do in this set of notes is Neutral. Definitions: Circles in Neutral Geometry circle is the set of points in a plane which lie at a positive, fixed distance r from some fixed point. The number r is called the radius and the fixed point is called the center of the circle. point P is said to be inside the circle, or be an interior point, whenever P < r. If P > r, then P is said to be outside the circle or to be an exterior point. The term radius also refers to any line segment joining point to any point on the circle. Theorem: line and a circle cannot have more than two points in common. ~ Suppose for contradiction that a line and a circle shared three points,,, and C. ecause they are on the circle, each of the points is equidistant from the center point. Thus = = C. WLG, assume **C. Then we have two isosceles triangles Δ and ΔC. Then p and p are base angles and each must be acute. Similarly, pc and pc must both be acute. ut this is a contradiction, since p and pc form a linear pair and are thus supplementary. This allows us to make some more: Definitions: line is tangent to or is a tangent line of the circle if it intersects the circle in exactly one point. the point of intersection is called the point of contact or point of tangency. line is a secant line for the circle if it intersects the circle in exactly two points.

2 Tangent Line Theorem: line is tangent to a circle iff it is perpendicular to the radius at the point of contact. ~ egin with a circle whose center is and radius is r, a point on the circle, and a line t passing through. Suppose line t is not perpendicular to. We show t cannot be a tangent. Drop a perpendicular to t. Find C on t such that **C and = C. Then ª and ªC are congruent by SS, and so C =, forcing C, a point of t, to be on the circle. So t is not tangent. C t Conversely, suppose t is perpendicular to. Let X be any other point on t. ªX is a right triangle with hypotenuse X. Thus X > = r, and X cannot lie on the circle. So t is tangent. X t

3 There are many terms associated with circles that we could define carefully, but we will trust you to be able to write definitions of these terms as needed, and simply illustrate them in a diagram: Chord Secant Radius Central ngle Diameter Point of Tangency or Contact Inscribed angle Tangent Semicircle ngle inscribed in semicircle Subtended or intersected chord or arc of ngle Diameter rc: Intersection of half plane and circle.

4 There are many theorems you can state and prove involving tangents, secants, and chords, including these: The center of a circle is the midpoint of any diameter. The perpendicular bisector of any chord of a circle passes through the center. line passing through the center of a circle and perpendicular to a chord bisects the chord. Two congruent central angles subtend congruent chords, and conversely. Two chords equidistant from the center of a circle have equal lengths, and conversely. tangent line contains no points interior to a circle. {, } lies outside the circle. If is a chord of a circle, then any point in lies inside the circle, and any point in Writing proofs for these is a good exercise, and we will leave these proofs mainly to the interested student (for whom Dancing with the Stars holds no more allure), and continue on to Elementary Circular Continuity, which is another theorem making full use of the deeper properties of the real numbers.

5 We will need a lemma; this is one I said earlier in the course that we would prove if we needed it: Lemma: Given ray and any point not on, define a function d(x) for any real x \$0 as the distance from to P on, where x is the distance from to P. That is,. Then d(x) is continuous. d( x) = Piff x= P ~ Pick ε > 0. Let R and S be such that R = x and S = y. Let y x = S R = RS < ε. y the Triangle Inequality, R S + RS and S R + RS. Either way, S R RS. Then we have d( y) d( x) = S R RS < ε. Thus d(x) is continuous. d(x) x P d(y) d(x) S R

6 Theorem (Elementary Circular Continuity): If a line l passes through an interior point of a circle (center and radius r), it is a secant of the circle, intersecting the circle in exactly two points. ~ Pick any other point on l and as in the lemma establish a function d( x) = Piff x= P for points on. Now since is interior to the circle and P = 0 if P =, we have = < r. Now find N on with N = 2r. y the triangle inequality, + N > N = 2r, so N > 2r -. Since < r, we have N > 2r - r or N > r. Thus, d(2 r) > r. ecause d(x) is a continuous function, < r and d(2 r) > r together imply that there is an x 0 between 0 and 2r with d( x0 ) = r(by the Intermediate Value Theorem from calculus). Find the point C on the ray with C = x 0. Then C = d( x0 ) = r, and so C lies on the circle. So C is a point of intersection of the line and the circle. d(x) d(2r) x P N Since a tangent line contains no interior points, it must be that l is a secant line, and we are done. However, we can actually find the other point of intersection without any difficulty.

7 nce we have C on the circle, it is easy to construct D on the circle as well: Drop a perpendicular from to a point H on line l. Find a point D with D * H * C and DH = CH. Then y SS, ÎHC ÎHD. y CPCF, D = C = r, so D is also an intersection point of the circle and line. D H C

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