Circles
Circles and
and ⇡r
r
Let
We’ll write
write CC to
to denote
denote the
the set
set ofof all
all vectors
plane
Let rr >> 0.0. We’ll
vectors inin the
the plane
r
whose
norm equals
equals r.r. That
That is,
is, CC isis the
the circle
circle ofofradius
radius rr centered
centered atat the
the
whose norm
point
0).
point (0,
(0,0).
C
r
r
Proposition
Proposition (3).
(3). The
The circle
circle CC’ isis aaconic.
conic. It’s
It’sthe
theset
set ofofsolutions
solutionsofofthe
the
quadratic
equation
quadratic equation
2
2
2
x2
++yy
2==r
Proof:
that
Proof: Recall
Recall
that the
the definition
definitionofofthe
thenorm
normofofaavector
vectorisis
p
2
2
||(x,
x ++yy
is the distance
(x,(0,0).
y) and (0, 0).
the distance
betweenbetween
(x, y) and
,, which
2
y)j == x
I(x,y)||
2
2
2
Also
notice that
that xx
2++yy
never negative,
negative, and
andthat
that rr isisalso
alsonot
Also notice
not negative.
negative.
2 isis never
Because
Because the
the square-root
square-root function
functionand
andthe
thesquaring
squaring_function
areinverse
inversefuncfunc
p function are
2
2
tions
x ++yy
tions for
for nonnegative
nonnegative numbers,
numbers, the
the equation
equatioii \/x2
equivalent by
by
2 == rr isis equivalent
2
2
2
invertible
function
to
the
equation
x
+
y
=
r
(just
square
both
sides
of
the
invertible function to the equation x
2+y
2 (just square
2 =r
pboth sides of the
2
former
x2 ++yy
equationto
to obtain
obtainthe
thelatter
latterequation).
equation). Therefore,
Therefore, \,/x2
former equation
and
2==rrand
2
2
2
xx
+
y
=
r
have
the
same
set
of
solutions.
2+y
2 have the same set of solutions.
2 =r
Putting
everything
together, we
Putting everything together,
we have
have
r
CCr={(x,y)
= { (x, y) 2ER
R2 | I ||(x,
y)|| ==r}
r}
2
pl(x,y)H
=={(x,y)ER
{ (x, y) 2 R2 | x2+y2=r}
x2 + y 2 = r }
2
2
2
2
2
=={{(x,
y) 2ERR
(x,y)
2| x12++yy
2}}
2==rr
⌅
Example.
Example.
•. The
The circle
circle ofofradius
radius 33 centered
centered atat (0,
theset
set ofofsolutions
solutionsofofthe
the
(0,0)0) isisthe
2
2
quadratic
equation
x
+
y
=
9.
quadratic equation x2 + y
2 =
Circles
Circles centered
centered at
at any
any point
point
We
Wecan
canuse
useProposition
Proposition33to
todescribe
describethe
theequation
equationofofany
anycircle
circleininthe
theplane.
plane.
186
170
Corollary (4). The circle of radius r centered at the point (a, b) 2 R2 is
a conic. It’s the set of solutions of the equation
(x
a)2 + (y
b)2 = r2
>
•1
Proof: C r is the circle of radius r centered at (0, 0). The planar transformation A(a,b) : R2 ! R2 move points horizontally by a and vertically by b.
Thus, A(a,b) (C r ) is the circle of radius r centered at the point (a, b).
S
Using POTS, the equation for A(a,b) (C r ) is given by precomposing the equa1
tion for C r , x2 + y 2 = r2 , with the function A(a,b)
= A( a, b) – the function
that replaces x with x a and y with y b. The resulting equation is
(x
a)2 + (y
b)2 = r2
⌅
Example.
• The circle of radius 5 centered at ( 2, 4) is the set of solutions of the
quadratic equation (x + 2)2 + (y 4)2 = 25.
*
*
*
*
*
*
*
*
*
*
*
*
*
The Unit circle
C 1 is called the unit circle. It’s the circle of radius 1 centered at the point
(0, 0). Half of the unit circle is above the x-axis, and half is below the x-axis.
\,J
0.
187
Some of the points in the unit circle include (1, 0), (0, 1), ( 1, 0), and
(0, 1). These are vectors of norm 1. The point p12 , p12 is also a vector of
norm 1
s
⇣ 1
⌘
⇣ 1 ⌘2 ⇣ 1 ⌘2 r 1 1 p
1
p , p
p
=
+ p
=
+ = 1=1
2 2
2
2
2
2
so p12 , p12 is also on the unit circle C 1 . It’s one of two points that are on
p1 , p1 .
the unit circle and on the line y = x. The other is
2
2
✓
◆
1 0
We can flip the circle over the y-axis using the matrix
. This is
0 1
the matrix that assigns the vector (x, y) to the vector ( x, y), which shows
p1 , p1
us two more points on the circle:
and p12 , p12 .
2
2
ID
(4,
(o’H)
(Oil)
I
(:t)
188
The Definition of
The
Definition
oftop
⇡ half of the unit circle.
length of the
r is the
⇡ is the length of the top half of the unit circle.
Because ir is the length of half of the circle, the length around the entire
Because
⇡ isthe
thecircumference
length of halfofofthe
thecircle
circle, equals
the length
around
theofentire
circle
called
277. The
length
the
circle – of
called
the circumference
of the (1,
circle
– equals
Thequarter
lengthofofthe
the
portion
the circle
between the points
(0, 1) 2⇡.
is one
0) and
portion of the of
circle
points
(1, 0)
and=(0, 1) is one quarter of the
the between
circle, so the
length
(27r)
circumference
it has
1
circumference of the circle, so it has length 4 (2⇡) = ⇡2 .
—
The flip over the y = x line interchanges the portions of the circle between
Theand
flip over the andbetween
y = x line interchanges
the circle
between
(1,0)
andthe
1). Thus,ofthese
two portions
(0,portions
1
1
1
1
p ,p
(1, 0) and p , p2 , and
between
and (0, 1). Thus, these two portions
namely
half of
2the2 total length between (1,0) and (0,1),
. have the same2 length,
it
have
the same
length, namely half of the total length between (1, 0) and (0, 1),
or
22)
4⇡
1 ⇡
or 2 2 = 4 .
0
—
—
/
/
\
1
?
1
i’)
/9\r
/oe -e.i ‘1Iitie.
7
(%,o)
(x
C’
173
189
\
e
The Circle divided into eight equal segments
The picture below shows the unit circle divided into eight segments, each
of length π4 .
:p
(i’o)
4?
1 -)
(o’i)
•17
4?
IL?
L)
(t-’o)
It will probably help to refer back to this page when answering #11-18 in
the exercises for this chapter.
190
The Winding function
We define the function wind : R → C 1 as follows: We let wind(0) = (1, 0).
If θ > 0, then wind(θ) is the point on C 1 obtained by beginning at the point
(1, 0) and traveling counterclockwise around the unit circle a length of θ. If
θ < 0, then wind(θ) is the point obtained by beginning at (1, 0) and traveling
clockwise around C 1 a length of |θ|.
0<0
Examples.
.
• wind(π) is the point on C 1 obtained by beginning at the point (1, 0)
and traveling counterclockwise around the unit circle a length of π. That
would take us exactly half-way around the circle to the point (−1, 0). Therefore, wind(π) = (−1, 0).
iT
çN
(i,o)
wht(’7r):(-i,o)
CI
4’z.
Lw’
\
191
a
-
• wind π4 is the point on the unit circle obtained by traveling a length
of π4 from the point (1, 0) counterclockwise
around the unit
circle.
As dis1
1
π
1
1
cussed previously, that point is √2 , √2 . That is, wind 4 = √2 , √2 .
ft
I
%
4
7f
‘-)
‘j.)
(iso)
Cl
• Because − π2 is negative, wind − π2 is the point arrived at by traveling
a length of π2 = | − π2 |clockwise around the circle from the point (1, 0). Thus,
wind − π2 = 0, −1 .
p
(o’i)
192
Period
Periodofofthe
thewinding
windingfunction
function
Notice
0). That
0), and
Noticethat
Thatis,is,ififyou
startatatthe
thepoint
point (1,
(1,0),
and
thatwind(2π)
wind(2ir)==(1,
(1,0).
youstart
travel
walk around
around the
the full
full circumference
circumference of
of the
the circle,
circle, you’ll
you’ll be
be back
back where
where you
you
started,
started,atatthe
thepoint
point(0,
1).
(0,1).
More
generally,
if
θ
∈
R,
More generally, if 9
IR,then
thenwind(θ)
wind(9) isisaapoint
point on
onthe
the circle,
circle, and
and once
once
atatthe
point
wind(θ),
we
could
travel
the
entire
circumference
of
the
circle
the point wind(&), we could travel the entire circumference of the circle
again,
again,a alength
lengthofof2π,
277,and
andwe’d
we’dbe
beback
backatatthe
thepoint
pointwind(θ).
wind(9). That
That is,
is,
wind(θ
wind(O++2π)
271)==wind(θ)
wind(9)
Functions
Functionsthat
thatbehave
behavelike
likethis
thisare
aresaid
saidtotohave
haveaaperiod
periodofof2π.
2ir.
C
Notice
Noticethat
thatwind(θ
wind(9++4π)
4ir)isisthe
thepoint
pointobtained
obtainedby
bystarting
startingatatwind(θ)
wind(&) and
and
then
thentraveling
travelingaround
aroundthe
thecircle
circletwice,
twice,aalength
lengthofof2(2π)
2(2ir)= 4π.
471. That
Thatwould
wouldput
put
ususback
at
the
point
wind(θ)
again,
so
that
wind(θ+4π)
=
wind(θ).
Similarly,
back at the point wind(9) again, so that wind(O+471) wind(9). Similarly,
wind(θ
wind(O++6π)
3(271),and
andtraveling
travelingcounterclockwise
counterclockwise
671)= wind(θ)
wind(9)because
because6π
671==3(2π),
around
the
circle
three
times
brings
you
back
to
where
you
around the circle three times brings you back to where you started;
started; and
and
wind(θ
lengththe
of the
circle
clockwise
also
wind(9− 2π)
271)==wind(θ)
wind(9)because
becausetraveling
travelingthearound
circle
clockwise
also
brings
back
to where
started.
brings
started.
youyou
back
to where
youyou
—
193
176
Exercises
For #1-6, use Corollary (4) to write an equation for the circle with given
radius and center. The equation should have the form
(x ± a)2 + (y ± b)2 = c
1.) center: (0, 0); radius: 4
4.) center: (0, 0); radius: 3
2.) center: (2, 3); radius: 1
5.) center: (4, −2); radius:
Exercises
1
3
3.) center: (−5, 7); radius: 12
6.) center: (−5, −7); radius: 5
For #1-6, write an equation for the circle with given radius and cent
The equation should have the form
1
2
2
Let C be the unit circle, the set of solutions
of +
the(yequation
(x + a)
2
± b)
2 = cx + y = 1.
For #7-10, use POTS to give an equation for the following subsets of the
plane. Write them in the form ax2 + by 2 = 1.
4.) center: (0,0); radius: 3
1.) center: (0,0); radius: 4
2 0
7.)
(C 1 ), the unit circle scaled by 2 in the x-coordinate and by 5
0 5
2.) center: (2,3); radius: 1
5.) center: (4, —2); radius:
in the y-coordinate.
3.) center: (—5,7);
radius:
6.) center: (—5,—7);
radius: 5
Let C’ be the unit circle, the set of solutions of the equation x
2+y
2 =
For #7-10, give an equation for the following subsets of the plane. Wr
them in the form ax
2 + by
2 = 1.
()
7.)
(C’), the unit circle scaled by 2 in the x-coordinate arid by
7 0
8.)
(C 1 ), the unit circle scaled by 7 in the x-coordinate and by 3
0 3
in the y-coordinate.
in the y-coordinate.
3
8.)
()
170\f
co3)LC
(C’), the194unit circle scaled by 7 in the x-coordinate and by
in the y-coordinate.
of the
plane.3 Wr
give(0.0):
an equation
subsets
For 1.)
radius:
4.) center:
4 the following
#7-10,
radius: for
center:
(0.0);
2 + by
2 = 1.
them in the form ax
5.) center: (4, —2); radius:
2.) center: (2, 3); radius: 1
7.)
(C’), the unit circle scaled by 2 in tile x-coordinate and by
1 6.) center: (—5, —7); radius: 5
3.) center: (—5, 7); radius:
0
1
in the by
y-coordinate.
9.) 4
(C 1 ), the unit circle scaled
in
the
x-coordinate
and by 2
4
0 2
thebey-coordinate.
the unit circle, the set of solutions of tile equation x2 + y
2
LetinC’
()
For #7-10, give an equation for tile following subsets of the plane. Wr
2 + by
2 = 1.
them in the form ax
7.)
10.)
()
()
(C’), the unit circle scaled by 2
in tue x-coordinate
and by
in the y-coordinate.
1 0
1
(C8.)), the unit(C’),
circletilescaled
10 scaled
in the by
y-coordinate.
unit by
circle
7 in the x-coordinate and by
0 10
in the y-coordinate.
(c’)
—I
I
For #11-18, write the given point in the plane in the form (a, b) for some
scaled by 7 in the x-coordinate and by
thewith
circlequestions.
unit these
(C’),
numbers a and b. 8.)
Page 190 will
help
()
in the y-coordinate.
177
15.) wind
11.) wind(8π)
12.) wind
7π
4
13.) wind −
7π
4
37π
4
16.) wind −
17.) wind
14.) wind(−π)
13π
2
125π
4
18.) wind(−4π)
177
For #19-22, find the set of solutions of the given equations.
19.) (3x + 4)(2x − 1) = x(6x − 2)
20.) (2 loge (x) − 3)2 = 100
√
21.) ex−4 + 1 = 5
22.)
1
x2
4
x
=3
195