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Subsections

# 5 Circles and lines

## 5.1 Circles

A circle is the set of points at fixed distance from a given point, its center. The distance is called the radius of the circle .

The circle divides the plane into two regions: the inside, which is the set of points at distance less than from the center , and the outside, which consists of all points having distance from greater than . Note that every line segment from to a point on has the same length .

A line segment from to a point on is also called a radius; this should cause no confusion.

A line segment connecting two points of is called a chord, if the chord passes through the center, then it is called a diameter.

As above, we also use the word diameter to denote the length of a diameter of , that is, the number that is twice the radius.

Proposition 5.1   A line intersects a circle in at most two points.

Proof. Suppose we had three points, , and , of intersection of with .

We first take up the case that is a diameter. In this case, we would have at least two of the three points on the same side of on ; hence we can suppose that and both lie on the same side of . However, by the ruler axiom (Axiom 6), we must have since . This contradicts our assumption that and both lie on .

We next take up the case that is not a diameter. We can assume that lies between and on . Draw the line segments, , , . Then , and are triangles. In fact, since , they are isosceles triangles. Let be the measure of the base angles of triangle . Then it is also the measure of the base angle of , and so it is also the measure of the base angle of . Since the two base angles at add up to , we obtain that each of the three triangles have two right angles, which is impossible.

Proposition 5.2   Let be a chord of a circle with center . Then the perpendicular bisector of passes through .

Exercise 5.1   Prove the preceding proposition.

## 5.2 Central angles

While we have spent a fair amount of time determining when two angles have the same measure, we have not discussed explicitly calculating the measure of an angle, except in the case of an angle of measure and a right angle (measure ). We shall do so now.

First, we assume the well-known property that the circumference (that is, the arc length) of a circle of radius is .

Now let and be two points on a circle of radius 1 and center . The radii and make two angles (the inner'' and the outer'' angles); call them and . An angle such as whose vertex is at the center of the circle is called a central angle. Notice that , no matter where and lie on . If and are the endpoints of a diameter, they divide the circle into two arcs, each of length ; note also that the measure of the angles and are also . In other cases, the length of the arc subtended by the angle will be whatever fraction of that is of the entire circle. For example, if is a right angle, it will take up of the circle, and the corresponding arc length will be . We define the measure of the angle to be the corresponding arc length when that angle is the central angle of a circle of radius 1.

Theorem 5.3 (measure of inscribed angle is half the central angle)   Let , and be points on the circle of radius . Draw the chords and , and draw the radii, and . Let be the measure of the inscribed angle . Then the measure of the central angle is . (Here we mean the angle which subtends the arc not containing .)

Proof. Draw the line . This divides quadrilateral into two isosceles triangles. Let be the measure of the base angles of , and let be the measure of the base angles of Then the measure of the requisite central angle is given by

## 5.3 Circumscribed circles

The circle is circumscribed about if all three vertices of the triangle lie on the circle. In this case, we also say that the triangle is inscribed in the circle.

Note that another way to describe a circle circumscribed about a triangle is to say that it is the smallest circle for which every point inside the triangle is also inside the circle. In this view, the problem of circumscribing a circle becomes a minimization problem. A given triangle lies inside many circles, but the circumscribed circle is, in some sense, the smallest circle which lies outside the given triangle.

It is not immediately obvious that one can always solve this minimization problem, nor that the solution is unique.

Proposition 5.4 (Uniqueness of Circumscribed Circles)   There is at most one circle circumscribed about any triangle.

Proof. Suppose there are two circles and which are circumscribed about . Since points , , and lie on both circles, and are chords. By Prop. 5.2, the perpendicular bisectors of and both pass through the centers of and . Since these two distinct lines can intersect in at most one point, and share the same center . Since is a radius for both circles, they have the same center and radius, and hence are the same circle.

Theorem 5.5 (Existence of Circumscribed Circles)   Given , there is always exactly one circle circumscribed about it.

Proof. We need to show that the perpendicular bisectors of the sides of meet at a point, and that this point is equidistant from all three vertices. Then the requisite circle will have this point as its center , and the radius will be the length of . Uniqueness was shown in Prop. 5.4.

Let and be the midpoints of sides and respectively. Draw the perpendicular bisectors of and , and let be the point where these two lines meet (note that need not be inside the triangle). Draw the lines , and .

We cannot have both that and (since ), hence we can assume without loss of generality that . Then we have , angles and are both right angles, and of course, . Hence, by SAS. In particular, . If , then we have shown that , from which it follows that there is a circumscribed circle with center and radius .

If , then we repeat the above argument to show that , from which, as above, it follows that . Again, this shows that there is a circumscribed circle.

Corollary 5.6   In any triangle, the three perpendicular bisectors of the sides meet at a point.

Exercise 5.2   Explain why Theorem 5.5 implies this corollary.

Corollary 5.7 (Three Points Determine a Circle)   Given any three non-colinear points, there is exactly one circle which passes through all three of them.

Exercise 5.3   Explain why this corollary follows from Theorem 5.5.

## 5.4 Tangent lines and inscribed circles

A line that meets a circle in exactly one point is a tangent line to the circle at the point of intersection. Our first problem is to show that there is one and only one tangent line at each point of a circle.

Proposition 5.8   Let be a point on the circle , and let be the line through perpendicular to the radius at . Then is tangent to .

Proof. There are only three possibilities for : it either is disjoint from , which cannot be, as is a common point; or it is tangent to at ; or it meets at another point . If meets at then is a triangle, where is a right angle. Since and are both radii, . Hence is isosceles. Hence . We have constructed a triangle with two right angles, which cannot be; i.e., we have reached a contradiction.

Proposition 5.9   If is a line tangent to the circle at the point , then is perpendicular to the radius ending at .

Proof. We will prove the contrapositive: if is a line passing through , where is not perpendicular to the radius, then is not tangent to .

Draw the line segment from to , where is perpendicular to . Let be the point of intersection of and . On , mark off the distance from to some point , on the other side of from . Since is perpendicular to , . By SAS, , and so . Thus both and lie on , and intersects in two points. Thus, is not tangent to .

Corollary 5.10   Let be a point on the circle . Then there is exactly one line through tangent to .

Exercise 5.4   Prove this Corollary.

A circle is inscribed in if all three sides of the triangle are tangent to . One can view the inscribed circle as being the largest circle whose interior lies entirely inside the triangle. (Note that it is not quite correct to say that the circle lies entirely inside the triangle, because the triangle and the circle share three points.)

We start the search for the inscribed circle with the question of what it means for the circle to have two tangents which are not parallel.

Proposition 5.11   Let be a point outside the circle , and let and be tangents to emanating from . Then the line segment bisects the angle between and .

Proof. Let be the point where is tangent to , for . Draw the lines and . Observe that , and that, since radii are perpendicular to tangents, , and these are both right angles.

By SSA, . Hence .

From the above, we see that if there is an inscribed circle for , then its center lies at the point of intersection of the three angle bisectors, and its radius is the distance from this point to the three sides. Hence we have proven the following.

Corollary 5.12 (Inscribed circles are unique)   Every triangle has at most one inscribed circle.

Theorem 5.13   Every triangle has an inscribed circle.

Proof. Let be the point of intersection of the angle bisectors from and in . Let be the point where the orthogonal from meets ; let be the point where the orthogonal from meets ; and let be the point where the orthogonal from meets .

Observe that, by AAS, . Similarly, and .

We have shown that the perpendiculars from to the three sides all have equal length; call this length . then the circle centered at of radius is tangent to the three sides of exactly at the points , and .

This theorem gives another proof of the result of exercise 2.16.

Corollary 5.14   The three angle bisectors of a triangle meet at a point; this point is the center of the inscribed circle.

Exercise 5.5   Give a proof of this corollary using the above theorem.

Exercise 5.6   Let and be such that , , and . Prove that .

Exercise 5.7   Let and be points on the circle . Let be the line tangent to at and let be the line tangent to at . Prove that if and are parallel, then the line segment is a diameter of .

Next: 6 Some Amusements Up: MAT 200 Course Notes on Previous: 4 Lengths, areas and
Scott Sutherland 2002-12-18