The circle

Equation of a circle

Take an orthonormal system of coordinates with axes x and y.
Say point M(a,b) is the center of a circle C with radius r.

 
                P(x,y) is point of C
                        <=>
                     |P,M| = r
                        <=>
               _______________________
              V (x - a)2  + (y - b)2   = r
                        <=>
 
                (x - a)2  + (y - b)2 = r2            (1)

The last equation is the equation of the circle C.

The equation of the circle with center M(a,b) and radius r is

(x - a)2 + (y - b)2 = r2


Expanding (1) we have an equation of the form 

 
 
        x2  + y2  - 2 a x - 2 b y + c = 0             (2)

Thus, each circle has an equation of the form (2).

Theorem

Each equation of the form 

 
 
        x2  + y2  + 2 m x + 2 n y + c = 0             (3)
with
 
        m2  + n2 - c > 0

is the equation of a circle.


Proof:
We try to transform (3) to the form (1).

 
x2  + y2  + 2 m x + 2 n y + c + m2  + n2  =  m2  + n2
 
                <=>
 
        (x + m)2  + (y + n)2  = m2  + n2  - c
 
If  m2  + n2  - c > 0 then we can write  m2  + n2  - c = r2

The equation (3) can be written in the form 

 
        (x + m)2  + (y + n)2  = r2
 
So, it is the equation of the circle with center M(-m,-n)
                  ____________
                |  2    2
and radius r = \| m  + n  - c

Intersection of circle and line.

The coordinates of the intersection points of a circle and a line are the solutions of the system formed by the resp. equations. If that system has just one solution, the line is a tangent line of the circle.
Example: 

 
C :     (x - 9)2  + (y - 6)2  = 25    (4)
 
a :     y = -2 x + 14                   (5)
 
(5) in (4) gives
 
        (x - 9)2  + (14 - 2 x - 6)2  = 25
<=>
        -5 x2  + 50 x - 120 = 0
<=>
        x =  4  or x = 6
Then
        y = 6   or y = 2

The intersection points are (6,2) ; (4,6)

Intersection points of two circles

The coordinates of the intersection points of two circles are the solutions of the system formed by the resp. equations.
Example:

 
C1 :    (x - 2)2  + (y - 5)2  = 25    (6)
 
C2 :    (x - 6)2  + (y - 13)2  = 65   (7)
 
The equations are equivalent with
     /
     |  x2 + y2  - 4 x - 10 y + 4 = 0
     |
     \  x2  + y2  - 12 x - 26 y + 140 = 0
 
<=>
     /
     |  x2  + y2  - 4 x - 10 y + 4 = 0
     |
     \  16 y + 8 x - 136 = 0
<=>
     /
     |  x2  + y2  - 4 x - 10 y + 4 = 0
     |
     \  x = 17 - 2 y
<=>
        ...

The intersection points are (-1,9) and (7,5).

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
 

 

Tangent line in a point of the circle

Take the circle C with center M(a,b) and radius r. 

 
 
C:      (x - a)2  + (y - b)2 = r2
 
and point P(xo,yo).
 
                b - yo
MP has slope    ------
                a - xo
                              a - xo
The tangent line has slope  - ------
                              b - yo
 
The tangent line has equation
 
           a - xo
y - yo = - ------   (x - xo)
           b - yo

Power of a point

Power of a point with respect 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 with respect to C. This power is strictly positive if P is outside C, it is 0 for P on C, and it is srtictly 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) with respect 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 with respect 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 with respect to C1 = the power of P with respect 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 with respect 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.