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.

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• 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.

3

• 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 .

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• 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.

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• 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 .

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• 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, .

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• 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.

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• 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.

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• 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 .
  .

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• 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.

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• 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.

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• 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

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• 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
  

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• 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.

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• 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 .

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• 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.

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• 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),
  

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• 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.

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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.

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• 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.

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• 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.

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• 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.

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• 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

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• 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.

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• 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.

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• 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.

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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.

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• 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.

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• 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 .

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• 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.

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• 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.

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• 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.

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• 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.

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• 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.

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• 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.

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• 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.

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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)

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

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• (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.

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• 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 .

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• 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
  

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• 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)

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• 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