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, AB and CD are secants,  is a tangent, OC is a
radius, AB and CD are chords and CD 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, OA, OB, and OC all have length r
and so OA  OB  OC .
• Hence
OAC  OCA,
OAB  OBA ,
• and
OBC  OCB.
• Comparing the first and third congruences, we see that
.OBC  OAB.. But this contradicts either the exterior
angle theorem: SinceOBC is an exterior angle of OAB it
must be larger than OAB. 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 OA   .
• Proof. Since this is an “if and only if” theorem, we have
two statements to prove: If OA is perpendicular to  at A,
then  is a tangent, and if  is tangent to the circle at A,
then OA   .
• We first consider the case in which OA 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, OA  OB .
• But OAB is a right triangle with hypotenuse OB . 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 OA 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 OA is not perpendicular to  , we construct B on 
such that OB is perpendicular to  . Then we construct C
on  (Fig. 5.4) such that AB  BC. It is not hard to see that
. OBA  OBC By SAS. Thus, OA  OC . But since OA 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 OA 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
.PA and PB are tangent to C. Then PA  PB.
• Proof. Consider the triangles OAPand OBP. Each has OP
for one side,OA  OB by definition of a circle, and  A and
. B are each right angles. Hence, OAP  OBP by SSA
for right triangles. So PA  PB , as claimed.
A
P
O
B
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 C be circles with centers O and O
and with equal radii. Let A and B be points on C and A
and B  be points on C . Then AB  AB if and only if
. AOB  AOB.

• Proof. (Fig. 5.5) Assume AB  AB, since C and Chave
equal radii, OA  OA and OB  OB, then AOB  AOB
by SSS, and AOB and AOBare corresponding parts of
congruent triangles. Conversely, if we assume that
. AOB  AOB, then AOB  AOBby SAS, and here

.AB  AB 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 AB .
• If it is not clear from the context which arc is meant, it is
proper to add a point in between, such as APB or AQB .
• Assume, as in Fig. 5.6, that APB is the smaller of the two.
Then we will define the measure of then arc APB to be
that of AOB and the measure AQB to be 360  AOB . In
the spirit of the lemma we will say that two arcs ABand CD
are congruent, written AB  CD , if they have the same
measure and if they come from circles of equal radii (or
from the same circles). So AB  CD if and only if they have
the same measure and if the segments AB and CD 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 AQB
is inscribed in the circle and that it subtends arc AB , to
refer to AB , the arc APB 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 ABC subtends the
1
1
arc AC, then
B  AC  AOC
2
2
• Proof. There are three possible cases to consider: Either O
lies on a side of ABC, or it lies on the interior of ABC , or
it lies outside of ABC .
• First, let O be on a side of ABC : say O is on AB .
Consider the triangle OBC . Since it is isosceles, B  C.
Now
180  B  C  COB
 2B  COB
• Hence 2B  180  COB  COA . Therefore, B  1 AOC  1 AC
2
2
as claimed.
• Next, assume O is inside of ABC as in Fig. 5.8(a).
Connect B to O and extend to a point P on the circle so
that BP will be a diameter. We may now apply the
previous case to the anglesABP and PBC, since O is on
one side of each of them.
• Hence
ABP 
1
AP
2
PBC 
and
• Adding these two equations yields
ABP  PBC 
1
PC
2
1
( AP  PC )
2
• The left hand-side of this equation is ABC and the righthand side is 1 AC . The last case, in which O is outside
2
[see Fig. 5.8(b)], is similar. The only change is that now
.ABC  ABP  PBC .
• Here are some easy consequences of our theorem.
• Corollary. If BAC and BAC are each inscribed in a
circle and if each subtends the same arc BC , then A  A
1
• Proof. A  1 BC and A  2 BC .
2
• Corollary. An angle inscribed in a semicircle must be a
right angle.
• Proof. In this case the arc subtended is 180.
• 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 AB and CD
which intersect at a point E inside the circle, then
AE . EB = CE . ED.
• Proof. The angles A and D each subtend the arc BC
and therefore are congruent. Likewise, C and B are
congruent, for they each subtend AD. Hence ACE ~ DBE
So
AE CE
DE

BE
• Cross-multiplication now yields the desired result.
• In the situation of the theorem and Fig. 5.9. we can also
calculate AEC .
• Theorem. Given intersecting chords in a circle
(as in Fig. 5.9),
AEC 
1
( AC  BD )
2
• Since the sum of the angles of AEC is 180 , we conclude
that AEC  180  A  C

• Now A  12 CB and C  12 AD . Also, we can express 180 in
terms of arcs: 180  1 .360  1 ( AC  BC  BD  AD)
2
2
• Substituting all of this into the equation for AEC yields


AEC  180  A  C
as claimed.

1
1
1
( AC  BC  BD  AD)  BC  AD
2
2
2

1
( AC  BD )
2
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 PO and find the midpoint M. Draw a
circle with diameter PO by making M the center and
MO = MP the radius. This circle will intersect C at two
points, A and B. Both PA and PB are solutions in that
both are tangent to C.
• Proof. How do we prove that PA and PBare tangent to C?
Recall that, according to a theorem in Section 5.1, we
simply need to show that OA is perpendicular to PAand OB
is perpendicular to PB . Now the angleOAP is inscribed
in the semicircle OAP with center M, and the angle OBP
is inscribed in the semicircle OBP . 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 x 2  ab. 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 AB
of length a + b, with intermediate point E such that AE = a
and EB = b. Next construct the midpoint M of AB . Using
M as center, we can now draw a circle with center M and
with AB as diameter. Finally, construct a line
perpendicular to AB through E. This line will intersect the
circle at points C and D. Then EC = ED = x.
• Proof. Since AB and CD are chords intersecting at E,
CE . ED = AE . EB = ab. So all we need to show is that
CE = ED. Draw MC and MD and consider the triangles

. MCE and MDE. Since MC and MD are radii, MC  MD,
and it is obvious that ME  ME . Also, CEM and DEM
are right angles. So, by SSA for right triangles, MCE  MDE
and CE and ED 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 x 2  7 x  12  0 ?
• 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 x 2  7 x  12 would be to factor it.
We write x 2  7 x  12  ( x  ?)( x  ?) 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 b 2
This corresponds to solving the quadratic equation
x 2  ax  b 2  0.
• Problem. Given line segments of lengths a and b, find two
lengths whose sum is a and whose product is b 2 . (Or,
construct a rectangle with a given area and a given
semiperimeter.)
• Solution. Let AB be a line segment of length a, and
construct a circle as in Fig. 5.12 with AB as diameter.
Construct a line  perpendicular to AB , 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 AB , meeting AB
at E. Then AE and EB 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 DE intersect the circle at the second
point F. Since AB and DF are intersecting chords, we
know that AE . EB = DE . EF. To complete the proof we
will show that DE = EF = b. DE 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 OD and OF . The triangles ODE and OFE
are congruent by SSA for right triangles.
Hence FE = DE = b and the proof is complete.
Application to Queen Dido’s
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 Dido’s 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 America’s 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 ABC 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 C  90 .
• Lemma 2. Let AB be fixed segment and let S = the set of
all points C such that ACB  90. Then S is a semicircle
with diameter AB .
• We will assume that the two lemmas are true and use them
to prove that the semicircle region is the solution to Queen
Dido’s problem. Then we will go back and prove the two
lemmas.
• Proof of Queen Dido’s Theorem. Let us call the region
that maximizes the area R and assume that R touches the
seashore along the line segment AB . Let C be any point
on the rest of the boundary of R.
• We first claim that the triangle ABC is contained in R.
The proof of this fact is by contradiction: If the boundary
of R sagged in and crossed AC or BC , 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 ACB must be a right angle. Again,
this proof will be by contradiction. The triangle ABC
divides R into three regions (Fig. 5.14) we have labeled 1,
2, and 3. If C  90 then we could produce a region with
the same perimeter and greater area using lemma 1. If C  90

we could push A and B further apart to make C  90 . This
“pushing” would not affect the lengths of AC and BC , 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 C is not less than a right angle. We can reason to a

similar contradiction if we assumed C  90. This
forces .

, as we90claimed.
C
• 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 AC (the dry side) such that C  90. 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 Dido’s
problem has a solution , then the solution is given by a
semicircle.
• We now backtrack and provide proofs of the lemmas.
1
ab
2 .
• Proof of Lemma 1. If C  90 then ABC has area
We will show that if C  90then ABC has area less than
. If we1 ab
take
as the BC
base, then
has area
.
,
ABC
2
1
where
h is the length of the altitude
. But
is AD
a leg of
AD
ah
2
the right triangle
, which
. So
AC

ADC has hypotenuse
1
1
or AD . AC
This implies
, as claimed.
ah  ab
h  b the area

2
2
• Proof of Lemma 2. We defined S to be the set
Let us now define to be the semicircle with {diameter
C ACB  90 }
and on the appropriate
S  side.
AB
• With this notation we need to show that S  S  . To show
that two sets are equal we need to show first that if C
belongs to S  , then it belongs to S; then we must show that
if C belongs to S, then it belongs to S  .
• If C belongs to S  then C is a point on the semicircle with
diameter AB . This means that ACB 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 ACB  90and we want to show
that C is on the semicircle with diameter AB . Our proof
will be by contradiction. If C is not on this semicircle, then
there is another point C where AC meets the semicircle.
As before, ACBis a right angle because it is inscribed in
a semicircle. Now we can get a contradiction if we
consider BCC. This triangle has two right angles, at C
and at C . This is impossible, and this completes the
proof.
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
.BC and DE intersect at a point A outside of the circle (as in
Fig. 5.17), then
• (a) A 
1
(CE  BD )
2
• (b) AB. AC  AD. AE
• Proof. (a): Consider ACE. A  180  C  E.
1
1
But C  BDE  ( BD  DE ) and E  1 CBD  1 (CB  BD ).
2
2
2
2
1
1


Also, 180  .360  [CB  BD  DE  CD].
2
2
• Now, by substitution,
1
1
1
1
A  [CB  BD  DE  CE ]  [ BD  DE ]  [CB  BD ]  [CE  BD ]
2
2
2
2
• (b) For this half of the theorem we consider the triangles ACD
and AEB (See Fig. 5.18). Each has  A for one angle.
Also, C  Esince each subtends the arc BD . Thus, by
AA, ACD ~ AEB . Hence AC  AD . If we cross
AE AB
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 AB and a tangent AC . Then BAC  1 AB.
2
• We remark that the line segment AB and the line AC make
two different angles with each other (they are
supplementary) and that there are two arcs that could be
labeled AB. Each angle is half of the arc AB 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 BACis less than or
equal to 90, as in the diagram. Let O be the center of the
circle and extend AO to a diameter AA .
• Then AA is perpendicular to AC and BAC  90  BA A.
1
1
1





BA
A

B
A
90

.
180

( AB  BA).
• But
and
.
2
2
2
The result now follows by subtraction. The proof in the
case of BAC obtuse is the same, except that BAC will
be 90  BAA rather than 90  BA A.
• 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 CD intersect at a point A,
then
(a) A  1 ( BD  BC )
2
2
(b) AB  AC. AD
• 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 BQC  BPC .
Then
1
A 
2
( BQC  BPC ).
B
A
Q
P
C