Circles

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

( )x h 2 ( )y k 2 R

( )x h 2 ( )y k 2 R2

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)

Examples:

Q. 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

( )x h 2 ( )y k 2 R2

( )x 2 2 ( )y 5 2 62

Q. What is the (standard) equation of the circle with radius 3 centered at (5,-8)?

ANS: h = 5, k = -8, R = 3 so

( )x 5 2 ( )y -8 2 32or

( )x 5 2 ( )y 8 2 9

Q. What are the center and radius of the circle with the following equation?

( )x 3 2 ( )y 5 2 10

ANS: Looking at we have

h = 3, k = 5,

( )x h 2 ( )y k 2 R2

R 10

Q. What are the center and radius of the circle with equation

ANS: compare to and we see that

h = -7, y = 0, and R = 11.

( )x 7 2 y2 121

( )x h 2 ( )y k 2 R2

( )x h 2 ( )y k 2 R2

Equations for circles may not be in standard form

Expand to get an equivalent equation:

x2 2 x h h2 y2 2 y k k2 R2

x2 y2 2 x h 2 k y h2 k2 R2

Which might also be written as

x2 y2 2 x h 2 k y h2 k2 R2 0

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.

x2 y2 2 x h 2 k y h2 k2 R2 0

Q. What is are the center and radius of the circle with the following equation?

x2 10 x 26 y2 14 y 0

x2 10 x y2 14 y 26 0

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 = 7

constant term = -26 = h2 k2 R2

But we just determines h an k so -26 = 25 + 49 - R2

So = 100 or R = 10. R2

Q. What is the standard form of the equation of the circle with the following equation?

x2 6 x 15 y2 2 y 0

Looking at x2 2 x h h2 y2 2 y k k2 R2 0

And comparing coefficients we see that -6 = -2h, 2 = -2k, and

= - 15

So h = 3, k = -2, and 9 + 1 – R^2 = -15 so 25 = R^2 and R = 5.

With this the standard equation is

h2 k2 R2

( )x 3 2 ( )y 1 2 52

Remark: The equation may be multiplied by a number. For instance if we had been

given we notice that the coefficients of x^2 and y^2 areboth 2. We want those coefficients to be 1 so we divide through by 2 to get

which we solved above.

2 x2 12 x 30 2 y2 4 y 0

x2 6 x 15 y2 2 y 0

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

( )x a1 2 ( )y a2 2 R12

( )x b1 2 ( )y b2 2 R22

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.

( )x a1 2 ( )y a2 2 R12_

( )x b1 2 ( )y b2 2 R22____________________________

( ) 2 a1 2 b1 x ( ) 2 a2 2 b2 y R12 a12 b12 a22 b22 R22 0

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.

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:

Subtracting 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 = -x+3/2 to get the points and

( )x 1 2 ( )y 2 2 4

x2 ( )y 1 2 3

2 x 4 2 y 1

x2

x

32

1

2

3 2 x2 x14

3

,14

234

14

234

,1

4234

54

234

,1

4234

54

234

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.

x2 y2 1

x2 1 1

x2 ( )y 3 2 22

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 -4*10*24 = -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

don’t 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.

x2 y2 1 ( )x 3 2 ( )y 1 2 1

6 x 10 2 y 0

y 3 x 5 x2 ( ) 3 x 5 2 1

10 x2 30 x 24 0

b2 4 a c

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 is:

“The 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 is

m -> ( , )

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) - >

Length R + a

Length bR ( )1 m2

1 m2

2 R m

1 m2

ba R

There is no real number to correspond to the point (-R,0) since the line that meets the circle“twice” at (-R,0) is vertical and has an infinite slope.For this reason we say that (-R,0) corresponds toinfinity

Q. 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 -> ( , ) = ( -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)) = 1/2

Note also that if one substitutes m = ½ as was done for m = -5 above thepoint returned is (9/5,12/5)/ ,

3 ( )1 ( )-5 2

1 ( )-5 2

3 *2 (-5)

1 ( )-5 2

Pythagorean triples:The 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 3^2 + 4^2 = 5^2 and(7,12,13) since 7^2 + 12^2 = 13^2

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 a*b 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

a2 b2 c2

( )a b 2 4 a b ( )a b 2

a-b2

2

a b

a+b2

2

,a s2 b t2 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 doesn’t 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 > 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 s=101 (and t=1) then the formula gives us y = 101, x = 5100, z = 5101

, ,xs2 t2

2y s t z

s2 t2

2

Pythagorean triples and the unit circle.

The unit circle is the circle of radius 1 centered at the origin. It has equation

x2 y2 1A 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

ac

2

bc

2

1 multiply through by c^2 to get a2 b2 c2

That is a,b,c is a pythagorean triple.

Conversely if a,b,c is a pythagorean triple then from We can divide through by c^2 to get

a2 b2 c2

ac

2

bc

2

1

That is (a,b,c) is a pythagorean triple if and only if (a/b,a/c) is a rational pointon the unit circle.

Arc Length Parameterization of the Circle: Radian Measure

B

O Ar

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 =

s

arc lengthradius

sr

Examples

A BO A=B, (angle = 0 degrees), s = 0 so s/r = 0,

radian measure = 0

A

B

O

AOB = 90 degrees = ¼ circle. Length of whole circle = so s length of ¼ circle =

So 90 degrees =

2 r2 r

4

r2

sr

2

2rad

AB O 180 degrees = ½ circle so arc has length

So for a half circle

rs

r

180 degrees rad

Conversion

180 degrees rad

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 *

180

degreeradian

1

180radian

degree1

radian180 degree =

230

180rad

Problem: 1.9 radian is how many degrees?

Sln: 1.9 radian* 1 = 1.9 radian * = deg180 degree radian

1.9 (180)

(r,0)O

(a, b)

Sin(t) and Cos(t)

t

r

( )sin tar

( )cos tbr

aa

b

r a2 b2

( )cos tb

a2 b2

( )sin ta

a2 b2

( )sin t 2 ( )cos t 2 1

Sin> 0cos>0Sin<0

Cos > 0

AO

B = (-2,7)

t

( )sin t-2

(-2)2 (7)2( )cos t

7

(-2)2 (7)2

Acute Angles

A

B

O

r

b

a

t

( )sin tar

( )sin topposite

hypotenuse

( )cos tbr

( )cos tadjacent

hypotenuse

t

(7,3)

(-2, -4)

3 – (-4) = 7

7 – (-2) = 9

92 72

t

What is the sin of the angle t at which the line meets the x-axis?

( )sin t7

92 72

( )cos t9

92 72