Circles

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Circles
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The circle with center C and radius R is all
points P that are exactly R from C If C
(h,k) and P (x,y) then Distance (P,C) R
Since we always prefer to avoid radicals if
possible we square both sides
This is the standard form of the equation for the
circle of radius R with center at (h,k)
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ExamplesQ. What is the standard form of the
equation of the circle with center (2,5) and
radius 6?Sln From h 2, k 5, R 6 so
the answer is
Q. What is the (standard) equation of the
circle with radius 3 centered at (5,-8)?ANS
h 5, k -8, R 3 so
or
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Q. What are the center and radius of the circle
with the following equation?
ANS Looking at
we have h 3, k 5,
Q. What are the center and radius of the circle
with equation
ANS compare to
and we see thath -7, y 0, and
R 11.
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Equations for circles may not be in standard
form
Expand to get an equivalent equation
Which might also be written as
Note that the coefficients of x and y are both 1
(in particular they are equal) and that the h
and k are contained in the coefficients of x
and y.
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Q. What is are the center and radius of the
circle with the following equation?
Looking at the form
And arranging the given equation to correspond
Then comparing coefficients we see
thatcoefficient of x -2h -10 so h
5coefficient of y -2k -14 so k
7constant term -26
But we just determines h an k so -26 25
49 -
So 100 or R 10.
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Q. What is the standard form of the equation of
the circle with the following equation?
Looking at
And comparing coefficients we see that -6 -2h,
2 -2k, and - 15 So h
3, k -2, and 9 1 R2 -15 so 25
R2 and R 5. With this the standard equation is

Remark The equation may be multiplied by a
number. For instance if we had beengiven
we notice
that the coefficients of x2 and y2 areboth
2. We want those coefficients to be 1 so we
divide through by 2 to get
which we solved
above.
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Points of intersection of two circles. Two
Circles can meet in 0, 1, or 2 points C and D
meet at no points, A and B at 1 point, B and C at
2 points. Q. What are the points of intersection
of the circle withradius R1 centered at A
(a1,a2) and the circle with radius R2 centered
at B (b1,b2)? Sln This is equivalent to
solving the two equationssimultaneously
The coordinates (p,q) of P1 will be solutions to
eachequation, similarly the coordinates of P2
will solve each.
Critical Observation If (p,q) is a solution to
each ofthe above equations it is a solution to
the difference of the two equations.
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_
____________________________
The important thing is that the difference of the
standard equations for twocircles is a linear
equation. Since the coordinates of P1 and P2
are solutions to each equation they are
solutions to the linear equation. Since two
points determine a line the difference is the
equation of the line through the points
ofintersection of the circles.
Thus the points of intersection of the two
circles are the points of intersection of the
line and either circle.
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Q. What are the points of intersection of the
circle of radius 2 centered at (1,2) and the
circle of radius sqrt(3) centered at
(0,1)?Sln The two equations are Subtractin
g the two equations and simplifying we get
or
y -x 3/2 Substitute for y in the second
equation (we could use either) we get
or
solving with the
quadratic formula gives x values of
from which we
substitute in to into y -x3/2 to get the
points
and


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Q. What are the points of intersection of the
circle of radius 1 centered at the origin and the
circle of radius 2 centered at (0,3)?Sln The
two equations are
and The difference of the two
equations gives y 1 Substituting for y in
the first equation we get
so we have onlyone solution x 0 which gives
us only one point (0,1)In this case the two
circles are tangent and have a single point in
common.
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Q. What are the points of intersection of the
circle with equations
and Sln. The difference
of the two equations is Solve for y and
substitute into the first equation to get
The quadratic
expands to When we apply the quadratic formula
the discriminant Is (-30)2 -41024 -60
which has no real square roots. This says that
thereare no real values for the x-coordinates of
the points of intersection there are nopoints
of intersection.The diagram shows the two
non-intersecting circles. Note that even though
the circles dont intersect the linear equation
defined by the difference of the two equations
stillmakes sense and is the equation of a line
which is perpendicular to the line connecting
centers of the two circles.
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Rational Parameterization of the
circle.Parameterization of a line gives us a
way to identify each point on the line witha
number in a one-to-one fashion. Parameterization
of the circle has the same objective to
identify each point on the circle with a number
Line with slope m through (-R,0)
Consider the circle of radius R centeredat the
origin. The parameterization isThe number m
corresponds to the other point ofintersection of
the line with slope m through thepoint (-R,0).
Since we can calculate the points of
intersection of a line and a circle we calculate
the correspondence ism -gt (
, ) Going
the other way is easy. If P (a,b) then the m
is the slope of the line from (a,b) to (-R,0)
so (a,b) - gt
Length b
Length R a
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There is no real number to correspond to the
point (-R,0) since the line that meets the
circletwice at (-R,0) is vertical and has an
infinite slope.For this reason we say that
(-R,0) corresponds toinfinity

• In the parameterization of the circle of radius
  3, centered at the origin by slopes of lines
  through (-3,0) which point on the circle
  corresponds to slope m -5?ANS -5 -gt (
  , )
  ( -36/13, -15/13)
• Q. What is the value of m that corresponds to
  the point P (9/5, 12/5) ?ANS m
  (12/5)/(3 (9/5)) 1/2Note also that if one
  substitutes m ½ as was done for m -5 above
  thepoint returned is (9/5,12/5)/
  ,

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Pythagorean triplesThe three integers (a,b,c)
are a pythagorean triple if One geometric
interpretation (for positive a,b,c) is that
these are the lengths of the right triangles
whose sides all have whole number length.Some
examples are (3,4,5) since 32 42 52
and(7,12,13) since 72 122 132 We can
make up formulas to generate pythagorean triples
based on the fact that for any numbers a and
b Divide by 4 This says that if ab is a
square and both a b and a-b are even then the
result is a pythagorean triple. We can do this
by takinga and b to both be squares and both be
odd or even. That is take
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With s and t both even or both odd and
is a pythagorean triple
Example If s 3, t 1 then we get x 4, y
3, t 5 If s 5, t 1 we get x 12, y
5, z 13 If s 5, t 7 we get x -12, y
35, z 37 The -12 doesnt affect the fact
that the sums of the squares of the first two is
thethird. We could avoid the negative sign by
insisting that s gt t but since we are
interested in the squares we can just take the
absolute value and 12, 35, 37 is a
pythagorean triple Note that by taking t 1 and
s odd numbers we get an infinite number of
righttriangles with integer length sides such
that for any odd number you can make a right
triangle with integer lengths such that the
chosen number is the length of theleg of
smallest length and the hypotenuse is one larger
than the other leg. For instanceif s101 (and
t1) then the formula gives us y 101, x
5100, z 5101
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Pythagorean triples and the unit circle.The
unit circle is the circle of radius 1 centered at
the origin. It has equation
A point (x,y) on the unit circle is a rational
point if x and y are rational numbers. That is if
x and y are each the ratio of two
integers.(0,1), (1,0), (3/5, 4/5) are
rational points on the unit circle. Any two
rational numbers can be written with a common
denominator so if(x,y) is a rational point on
the unit circle then we can assume a a/c, y
b/cwith a,b,c integers. To say that (x,y) is
on the unit circle says that
multiply through by c2 to get
That is a,b,c is a pythagorean triple.
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Conversely if a,b,c is a pythagorean triple then
from We can divide through by c2 to get
That is (a,b,c) is a pythagorean triple if and
only if (a/b,a/c) is a rational pointon the
unit circle.
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Arc Length Parameterization of the Circle Radian
Measure
B
s
Angle AOB is defined by two rays OA and OB The
angle cuts an arc out of any circle with center
at the vertex, O. If the length of the arc is
s then s is some multiple of the length of the
radius. That is the arc is some number of
radiuses. This number of radiuses does not
depend on the radius different circles
centered at O give the same number of radiuses.
We use radian rather than radiuses Radian
measure of AOB
O
A
r
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Examples
A B
AB, (angle 0 degrees), s 0 so s/r 0,
radian measure 0
O
B
AOB 90 degrees ¼ circle. Length of whole
circle so s length of ¼ circle
So
90 degrees
A
O
rad
B
A
O
180 degrees ½ circle so arc has length
So for a half circle
rad
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Conversion
or
Problem 230 degrees is how many radians? Choose
the conversion factor that causes the old unit to
cancel Sln 230 degree 1 230 degree

rad
Problem 1.9 radian is how many degrees?Sln
1.9 radian 1 1.9 radian
deg

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Sin(t) and Cos(t)
a
a
r
t
b
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B (-2,7)
Singt 0cosgt0
Sinlt0Cos gt 0
t
A
O
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Acute Angles
B
r
a
t
O
b
A
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What is the sin of the angle t at which the line
meets the x-axis?
(7,3)
t
3 (-4) 7
t
(-2, -4)
7 (-2) 9