The Circle --- Extension




The Circle --- Basics

The basic concepts about the circle were treated on the page The Circle -- Basic concepts

Parametric equations of the circle

Let C be the curve with parametric equations [ x = a + r cos(t) , y = b + r sin(t) ].
 
     P(x,y) is on  C
<=>
    / x = a + r cos(t)
    \ y = b + r sin(t)

<=>
    / (x - a)/r = cos(t)
    \ (y - b)/r = sin(t)

<=>
     ( (x - a)/r )2 + ( (y - b)/r )2 = 1
<=>
     (x - a)2  + (y - b)2 = r2
<=>
    P(x,y) is on the circle with center (a,b) and radius r
Conclusion: The circle C with center (a,b) and radius r has parametric equations
[ x = a + r cos(t) , y = b + r sin(t) ].

With each t corresponds a point on the circle and vice versa.

P(a + r cos(t), b + r sin(t)) is a variable point on the circle C.

Example 1:

The circle C has an equation (x - 3)2 + (y - 6)2 = 25 and P(6,10) is on C. Say M is the center of C. Find the points A and B, on C, such that |PA|=|PB| = 5.

Since the radius of the circle is 5, the triangles MPA and MPB are equilateral. The angles are pi/3 radians.

The parametric equations of the circle are [ x = 3 + 5 cos(t) , y = 6 + 5 sin(t) ].

Point P has a t-value to such that cos(to)=0.6 and sin(to)= 0.8. From this we have to = 0.927295218

For the point A the t-value is to+pi/3 and then A = A(1.036 , 10.598)

For the point B the t-value is to-pi/3 and then B = B(7.964 , 5.402)

Example 2:

We take an arc of a circle with parametric equations
[ x = cos(t) , y = 1 + sin(t) ] with t in [- pi/2 , 0]
and the curve y = arccos(x).
Calculate the intersection point of these curves.

Therefore we have to calculate the t-value such that 1 + sin(t) = arccos(cos(t)).
Since t is in [- pi/2 , 0], arccos(cos(t)) = -t.
So, we calculate the t-value such that sin(t) + t + 1 = 0.
We can't solve the last equation algebraically. With a plot of sin(t) + t + 1, we can find the first approximation t = -0.51 to the root. With an efficient iteration method we find a very good approximation in only 4 steps.
-0.51
-0.51090446454
-0.510972899149
-0.510973425307
-0.510973429357

The requested intersection point is approximately ( 0.87 , 0.51)

 

         

Four points on a circle and equation of a circle through 3 points.

 
Given : points P1(x1,y1); P2(x2,y2); P3(x3,y3); P4(x4,y4)
        No set of three points are on one line.

        The four points are on one circle

                <=>

        There exist a circle

        x2  + y2  + a x + b y + c = 0
        such that the points are on that circle.

                <=>

        There are numbers a,b and c such that
     /
     |  x12  + y12  + a x1 + b y1 + c = 0
     |  x22  + y22  + a x2 + b y2 + c = 0
     |  x32 + y32   + a x3 + b y3 + c = 0
     |  x42  + y42  + a x4 + b y4 + c = 0
     \

                <=>

        There are numbers a,b and c such that

          /
          | a x1 + b y1 + c = -(x12  + y12 )
          | a x2 + b y2 + c = -(x22  + y22 )
          | a x3 + b y3 + c = -(x32 + y32  )
          | a x4 + b y4 + c = -(x42  + y42 )
          \


        This is a system of four equations with 3 unknowns.
        We know from the theory of systems of linear equations that
        The coefficient matrix is

     [   x1     y1   1   ]
     [   x2     y2   1   ]
     [   x3     y3   1   ]
     [   x4     y4   1   ]

        Because P1,P2,P3 are not on one line, the rank of this
        matrix is three. The last equation is the side equation.
        The characteristic determinant of this equation is


     |  x1     y1   1  -(x12  + y12 ) |
     |  x2     y2   1  -(x22  + y22 ) |
     |  x3     y3   1  -(x32  + y32 ) |
     |  x4     y4   1  -(x42  + y42 ) |


        This system has a solution for a,b and c if and only if
        this determinant is 0. From properties of determinants
        this condition is

     |(x12  + y12 )   x1     y1   1 |
     |(x22  + y22 )   x2     y2   1 |
     |(x32  + y32 )   x3     y3   1 |   = 0
     |(x42  + y42 )   x4     y4   1 |

 
The four points P1(x1,y1); P2(x2,y2); P3(x3,y3); P4(x4,y4)
are on one circle

                <=>

     |(x12  + y12 )   x1     y1   1 |
     |(x22  + y22 )   x2     y2   1 |
     |(x32  + y32 )   x3     y3   1 |   = 0
     |(x42  + y42 )   x4     y4   1 |


Corollary :
 
Given: P1,P2,P3 are not on one line

        Point P(x,y) is on the circle defined by P1,P2,P3

                        <=>
        P,P1,P2,P3 are on a circle

                        <=>
     |(x2  + y2   )   x      y    1 |
     |(x12  + y12 )   x1     y1   1 |
     |(x22  + y22 )   x2     y2   1 |
     |(x32  + y32 )   x3     y3   1 |   = 0

 
The equation of a circle through 3 given points
    P1(x1,y1); P2(x2,y2); P3(x3,y3) is

     |(x2  + y2   )   x      y    1 |
     |(x12  + y12 )   x1     y1   1 |
     |(x22  + y22 )   x2     y2   1 |
     |(x32  + y32 )   x3     y3   1 |   = 0


Power of a point

Power of a point relative to a circle

Take a point P and a circle C with center M and radius r.
The distance |P,M| = d.
A variable line through P intersects the circle in points A and A'.
Say MN is the perpendicular bisector of [A,A'].
 
       
Then we have (vectors in bold face)
 
        PA . PA' = (PN + NA)(PN + NA')
                 = (PN + NA)(PN - NA)
                     2     2
                 = PN  - NA

                 = |P,N|2  - |N,A|2

Now     |P,N|2 = d2  - |M,N|2   and |N,A|2 =   r2  - |M,N|2

Thus    PA . PA' = d2 - r2
This result depends only on the distance d and the radius r.
It is called the power of P relative to C. This power is strictly positive if P is outside C, it is 0 for P on C, and it is strictly negative for P inside C.

Analytic expression of the power of a point

The a circle C with equation
 
        (x - a)2  + (y - b)2 - r2  = 0
and a point P(xo,yo).
Now, the power of P is
 
        d2 - r2  = (xo - a)2  + (yo - b)2 - r2
 
The power of point P(xo,yo) relative to circle C with equation
        (x - a)2  + (y - b)2 - r2  = 0
is
        (xo - a)2  + (yo - b)2 - r2

Example:
 
C :     (x - 2)2  + (y - 3)2  = 25  and P(3,1)

The power of P relative to C is  1 + 4 - 25 = - 20

radical axis

Take two circles
 
C1:     (x - a)2  + (y - b)2 - r2  = 0
C2:     (x - c)2  + (y - d)2 - r'2 = 0
We look for the set of all points P(x,y) such that
 
  The power of P relative to C1 = the power of P relative to C2.

                                <=>
  (x - a)2  + (y - b)2 - r2  =  (x - c)2  + (y - d)2 - r'2

                                <=>
                (a - c) x + (b - d) y + k = 0
From this, we see that the set is a line.
This line is called the radical axis of the circles.
The slope of that line is (a-c)/(d-b)
The central line connecting the centers has slope (d-b)/(c-a).
Hence, the radical axis is orthogonal with the central line of the circles.
Example:
 
C1:     x2  + y2  = 25

C2:     (x - 2)2  + (y - 3)2  = 9

The radical axis is the line 4 x + 6 y - 29 = 0

radical center

Take three circles C1, C2, C3.
If the radical axis of C1 and C2 meets the the radical axis of C2 and C3 in point S, then S has the same power relative to the three circles.
The point S is called the radical center of C1, C2 and C3.
Then, of course, S is on the third radical axis too.

Exercises

The given solution is not 'the' solution.
Most exercises can be solved in different ways.
It is strongly recommended that you at least try to solve the problem before you read the hidden solution.




Topics and Problems

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