Chapter 5 : Circles

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Chapter 5 : Circles

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Title: Chapter 5 : Circles

1
Chapter 5 Circles

A circle is defined by a center and a radius.
Given a point O and a length r, the circle with
radius r and center O is defined to be the set of
all points A such that OA r. Much of this
chapter will be concentrated with the
relationships among circles and various lines.
The lines commonly associated with circles are as
follows
A secant is a line that intersects a circle at
two points.
A tangent is a line that intersects a circle in
one point.

A radius is a line segment with one end point at
the center and the other on the circle. Note, as
before, that the word radius sometimes refers
to the length of the segment.
A chord is a line segment with both end points on
the circle.
A diameter is a chord that contains the center.
In Fig. 5.1, and are secants, is
a tangent, is a radius, and
are chords and is a diameter.

We first study intersections of circles and
lines.
Lemma. A circle and a line cannot intersect in
three or more points.
Proof. The proof will be by contradiction. Assume
that we have a circle with center O and radius r,
and a line with distinct points A, B, and C
on both the line and circle. By the definition of
circle, , and all have length
r and so .

Hence
and
Comparing the first and third congruences, we see
that . . But this
contradicts either the exterior angle theorem
Since is an exterior angle of
it must be larger than . This
contradiction proves the lemma.

Theorem. Let A be a point on a circle with center
O, and let be a line through A. Then
is tangent to the circle at A if and only if
.
Proof. Since this is an if and only if theorem,
we have two statements to prove If is
perpendicular to at A, then is a
tangent, and if is tangent to the circle at
A, then .

We first consider the case in which is
assumed to be perpendicular to , and we want
to show that is a tangent.
So, by way of contradiction we assume that
is not tangent to the circle and that there is
another point B on the intersection of the circle
with . Since A and B are both points on the
circle, .

But is a right triangle with
hypotenuse . This gives a contradiction
because the hypotenuse of a right triangle has to
be longer than either of the legs.
Next, to prove the converse, we assume that
is not perpendicular to and we will prove
that is not tangent to the circle. In order
to do this we will find another point in the
intersection of the circle with .
Since is not perpendicular to , we
construct B on such that is
perpendicular to . Then we construct C on
(Fig. 5.4) such that . It is not
hard to see that . By
SAS. Thus, . But since is a
radius, this implies that C is also a point on
the circle and so completes the proof.

This theorem has a number of consequences. First,
we can use it to construct tangents.
Problem. Given a circle C with center O and point
A on C, construct a line through A and
tangent to C.
Solution. Line will be the line perpendicular
to at A.
Corollary. If C is any circle and A is a point on
C, then there exists a unique line through A
tangent to C.
Both the construction and corollary are easy
consequences of the theorem and we omit the
proofs.

Theorem. Let C be a circle with center O and let
A and B be points on C. Assume P is a point
external to C and that . and
are tangent to C. Then .
Proof. Consider the triangles and
. Each has for one side, by
definition of a circle, and and . are
each right angles. Hence,
by SSA for right triangles. So ,
as claimed.

A
P
O
B
10
Arcs and Angles

In this section we discuss circular areas and how
they are measured. We then prove a theorem
relating sizes of arcs to angles inscribed in
them. This will prove to be a key result for the
rest of our study of circles and also for later
chapters. We begin with an easy lemma.
Lemma. Let C and be circles with centers O
and and with equal radii. Let A and B be
points on C and and be points on
. Then if and only if .
.

Proof. (Fig. 5.5) Assume , since C
and have equal radii, and
, then by
SSS, and and are
corresponding parts of congruent triangles.
Conversely, if we assume that .
, then
by SAS, and here .
are corresponding parts of congruent
triangles.

Now let C be any circle and A and B be two points
on C. A and B divide the circle into two arcs,
and, confusingly enough, both arcs are referred
to as .
If it is not clear from the context which arc is
meant, it is proper to add a point in between,
such as or .
Assume, as in Fig. 5.6, that is the
smaller of the two. Then we will define the
measure of then arc to be that of
and the measure to be
. In the spirit of the lemma we will say that
two arcs and are congruent, written
, if they have the same measure and
if they come from circles of equal radii (or from
the same circles). So if and only
if they have the same measure and if the
segments and are congruent.

Moreover, in the spirit of Chapter 0, we will
often identify an arc with its degree measure.
Given three points on a circle, such as A,Q, and
B in Fig. 5.6, we will say that is
inscribed in the circle and that it subtends arc
, to refer to , the arc that
does not contain the vertex Q.
Note that in the statement of this theorem we use
our notation convention, which identifies angles
and arcs with the real numbers that are their
degree measures.
Theorem. If we are given a circle with center O
and containing points A, B, and C so that
subtends the arc , then

Proof. There are three possible cases to
consider Either O lies on a side of ,
or it lies on the interior of , or it
lies outside of .
First, let O be on a side of say O
is on . Consider the triangle .
Since it is isosceles, . Now

Hence .
Therefore, as claimed.
Next, assume O is inside of as in Fig.
5.8(a). Connect B to O and extend to a point P on
the circle so that will be a diameter. We
may now apply the previous case to the angles
and , since O is on one side of
each of them.

Hence
and
Adding these two equations yields
The left hand-side of this equation is
and the right-hand side is . The last
case, in which O is outside see Fig.
5.8(b), is similar. The only change is that now
. .
Here are some easy consequences of our theorem.
Corollary. If and are each
inscribed in a circle and if each subtends the
same arc , then
Proof. and .

Corollary. An angle inscribed in a semicircle
must be a right angle.
Proof. In this case the arc subtended is .
A third consequence of our theorem is the
following slightly more difficult but extremely
useful theorem.
Theorem. If we are given a circle with chords
and which intersect at a point E inside
the circle, then AE . EB CE . ED.

Proof. The angles and each subtend
the arc and therefore are congruent.
Likewise, and are congruent, for
they each subtend . Hence
So
Cross-multiplication now yields the desired
result.
In the situation of the theorem and Fig. 5.9. we
can also calculate .
Theorem. Given intersecting chords in a circle
(as in Fig. 5.9),

Since the sum of the angles of is
, we conclude that
Now and . Also, we
can express in terms of arcs
Substituting all of this into the equation for
yields
as claimed.

20
Applications to Constructions

We now show how these results can be applied to
solve some construction problems.
Problem. Given a circle C with center O and a
point P outside of C, find a line that
contains P and that is tangent to C.
Solution. Connect and find the midpoint M.
Draw a circle with diameter by making M
the center and MO MP the radius. This circle
will intersect C at two points, A and B. Both
and are solutions in that both are
tangent to C.

Proof. How do we prove that and are
tangent to C? Recall that, according to a theorem
in Section 5.1, we simply need to show that
is perpendicular to and is
perpendicular to . Now the angle
is inscribed in the semicircle with
center M, and the angle is inscribed in
the semicircle . Hence, each of them is
a right angle.

Problem. Given line segments of lengths a and b,
construct a line segment of length x so that
. In more geometrical terms, construct x
such that a square with side x would have area
equal to a rectangle with sides of length a and
b.

Solution. First, as in Fig. 5.11, construct a
line segment of length a b, with
intermediate point E such that AE a and EB b.
Next construct the midpoint M of . Using
M as center, we can now draw a circle with center
M and with as diameter. Finally,
construct a line perpendicular to through
E. This line will intersect the circle at points
C and D. Then EC ED x.
Proof. Since and are chords
intersecting at E, CE . ED AE . EB ab. So
all we need to show is that CE ED. Draw
and and consider the triangles
. and . Since and
are radii, , and it is
obvious that . Also,
and are right angles. So, by SSA for
right triangles, and
and are corresponding parts of congruent
triangles.

Our third construction is related to the problem
of solving a quadratic equation. How would you
solve an equation such as ?
You could use the quadratic formula and, in
principle, now that we know how to take square
roots geometrically, we could try to develop a
geometric construction based on the quadratic
formula.
Another method of solving would
be to factor it. We write
and try to find two numbers whose
sum is 7 and whose product is 12. Of course, 3
and 4 work and they would be the solutions. The
geometric problem we will try to solve is to find
two segments with sum a and product equal to a
given square This corresponds to solving the
quadratic equation

Problem. Given line segments of lengths a and b,
find two lengths whose sum is a and whose product
is . (Or, construct a rectangle with a given
area and a given semiperimeter.)
Solution. Let be a line segment of length
a, and construct a circle as in Fig. 5.12 with
as diameter. Construct a line
perpendicular to , at point A. Choose a
point C on such that AC b. Draw a line
through C, perpendicular to , and that meets
the circle at point D. Finally, drop a
perpendicular from D to , meeting at
E. Then and are the solutions to the
problem.

Proof. We need to calculate the sum and product
of AE and EB. First,
AE EB AB a.
As for the product, let intersect the
circle at the second point F. Since and
are intersecting chords, we know that AE . EB
DE . EF. To complete the proof we will show
that DE EF b. is a side of the
rectangle DEAC, so DE AC b.

To compute EF, let O be the center of the circle
and draw the radii and . The
triangles and are
congruent by SSA for right triangles.
Hence FE DE b and the proof
is complete.

28
Application to Queen Didos Problem

Queen Dido was promised as much land along the
coast as she could cover with an ox hide. In
order to get as much as possible, she cut and
sewed the hide and made it into a long rope.
According to the legend, this is how the ancient
city of Carthage was found.
From a mathematical point of view, here is the
problem Queen Dido faced She wanted to construct
a region such that one side would be a straight
line (the seashore), the rest of its boundary
would be a fixed length (the length of her rope)
and its area would be as great as possible.

Ancient state of North Africa, and at times also
the southwestern part of the Mediterranean basin,
lasting from about 9th century BCE to 146 BCE.
From the 8th century till the 3rd century BCE,
Carthage was the dominating power of the western
half of the Mediterranean.
The state had its name from the city of Carthage,
out on the coast, 10 km from today's Tunis,
Tunisia. Carthage had been founded in the 9th
century by Phoenician traders of Tyre. Carthage
had two first class harbours, and therefore an
advantage with the most efficient means of
communications of those days, the sea. The
Carthaginians soon developed high skills in the
building of ships and used this to dominate the
seas for centuries. The most important
merchandise was silver, lead, ivory and gold,
beds and bedding, simple, cheap pottery,
jewellery, glassware, wild animals from African,
fruit, nuts.

It turns out that the solution to Queen Didos
problem will be a bit vague on a few technical
points. For more details about this as well as
other interesting geometric optimization
problems, refer to Geometric Inequalities, by
Nicholas D. Kazarinoff, in the Mathematical
Association of Americas New Mathematical
Library, volume 4, 1961.
There are two ingredients to the proof, which we
will prove as separate lemmas.
Lemma 1. Of all triangles with BC
equal to a given length a and AC equal to a given
length b, the triangle of maximum area is the one
with .

Lemma 2. Let be fixed segment and let S
the set of all points C such that
. Then S is a semicircle with diameter .
We will assume that the two lemmas are true and
use them to prove that the semicircle region is
the solution to Queen Didos problem. Then we
will go back and prove the two lemmas.
Proof of Queen Didos Theorem. Let us call the
region that maximizes the area R and assume that
R touches the seashore along the line segment
. Let C be any point on the rest of the
boundary of R.

We first claim that the triangle is
contained in R. The proof of this fact is by
contradiction If the boundary of R sagged in and
crossed or , then by pushing it
out we could produce a region with an equal
perimeter and greater area and this would
contradict our definition of R.

Next, we claim that must be a right
angle. Again, this proof will be by
contradiction. The triangle divides R
into three regions (Fig. 5.14) we have labeled 1,
2, and 3. If then we could produce
a region with the same perimeter and greater area
using lemma 1. If
we could push A and B further apart to make
. This pushing would not affect the
lengths of and , so we could still
fit regions 1 and 3 together with our new region
2.
In the new figure the area would be greater as
guaranteed by lemma 1, and the perimeter would be
the same. Since we assumed that R has maximum
area, this is impossible and so is not
less than a right angle. We can reason to a
similar contradiction if we assumed
. This forces . , as we claimed.

Now we are done, by use of lemma 2. Our region R
has a boundary away from shore that consists of
points C on a given side of (the dry side)
such that . Hence the boundary of R
is a semicircle.
Note that we omitted some technical details. We
assumed without proof that the problem has a
solution! This means that what we have really
proven is that if Queen Didos problem has a
solution , then the solution is given by a
semicircle.
We now backtrack and provide proofs of the
lemmas.

Proof of Lemma 1. If then
has area . We will show that if
then has area less than .
If we take as the base, then has
area . , where h is the length of
the altitude . But is a leg of the
right triangle , which has hypotenuse
. So or . This
implies the area , as claimed.
Proof of Lemma 2. We defined S to be the set
Let us now define to be the
semicircle with diameter and on the
appropriate side.

36
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37

With this notation we need to show that
. To show that two sets are equal we need to
show first that if C belongs to , then it
belongs to S then we must show that if C belongs
to S, then it belongs to .
If C belongs to then C is a point on the
semicircle with diameter . This means that
is inscribed in a semicircle, and we
know that an angle inscribed in a semicircle must
be a right angle. So, by definition, C will be an
element of S.

Conversely, now assume that C belongs to S. This
means that we are assuming that
and we want to show that C is on the semicircle
with diameter . Our proof will be by
contradiction. If C is not on this semicircle,
then there is another point where
meets the semicircle. As before, is a
right angle because it is inscribed in a
semicircle. Now we can get a contradiction if we
consider . This triangle has two right
angles, at C and at . This is impossible,
and this completes the proof.

39
More on Arcs ands Angles

We proved various theorems concerning chords
before. In Now, we will prove analogous theorems
for secants and tangents.
Theorem. If B, C, D, and E are points on a circle
such that . and intersect at a point A
outside of the circle (as in Fig. 5.17), then
(a)
(b)

Proof. (a) Consider .
But
and
Also,
Now, by substitution,

(b) For this half of the theorem we consider the
triangles and (See Fig. 5.18).
Each has for one angle. Also,
since each subtends the arc . Thus, by AA,
. Hence
. If we cross multiply we get (b).
We now turn to the case of tangents. In order to
repeat the proof from the secant case, we need
this preliminary result.

Theorem. Assume at a point A on the circle that
there is a chord and a tangent .
Then
We remark that the line segment and the
line make two different angles with each
other (they are supplementary) and that there are
two arcs that could be labeled . Each angle
is half of the arc that it cuts off-the
larger angle corresponds to the larger arc and
the smaller angle corresponds to the smaller arc.
Proof. Assume that C is such that is
less than or equal to , as in the diagram.
Let O be the center of the circle and extend
to a diameter .

Then is perpendicular to and
But and
. The result now
follows by subtraction. The proof in the case of
obtuse is the same, except that
will be rather than

With this tool, the following theorems can be
proved easily.
Theorem. If B, C, and D are points on a circle
such that the tangent at B and the secant
intersect at a point A, then
(a)
(b)

Theorem. Suppose B and C are points on a circle
such that the tangent at B and the tangent at C
meet at a point A. Let P and Q be points on the
circle such that . Then

B
A
Q
P
C

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