Next: 6 Some Amusements
Up: MAT 200 Course Notes on
Previous: 4 Lengths, areas and
Subsections
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.
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.
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
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.
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.
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