The circle

• Equation of a circle
  
• Inverse property
  
  • Examples
    
• Parametric equations of the circle
  
• Intersection of circle and line.
  
• Intersection points of two circles
  
• Four points on a circle and equation of a circle through 3 points.
  
• Tangent line in a point of the circle
  
• Power of a point
  
  • Power of a point with respect to a circle
    
  • Analytic expression of the power of a point
    
  • radical axis
    
  • radical center
    
• Exercises
  

Equation of a circle

 
                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)

 

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

Inverse property

 

        x2  + y2  + 2 m x + 2 n y + c = 0             (3)
with

        m2  + n2 - c > 0

 
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

 
        (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

Examples

• The equation x² + y² + 2 m x = 0 is the equation of a circle. The center is (-m,0) on the x-axis and the radius is |m|.
• The equation x² + y² + 2 n y = 0 is the equation of a circle. The center is (0,-n) on the y-axis and the radius is |n|.
• The equation x² + y² + 2 m x + 2 n y = 0 is the equation of a circle. The center is M(-m,-n) and O(0,0) is on the circle.
• If c is negative, the equation x² + y² + 2 m x + 2 n y + c = 0 is the equation of a circle.

Parametric equations of the circle

 
     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

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 t_o such that cos(t_o)=0.6 and sin(t_o)= 0.8. From this we have t_o = 0.927295218

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

For the point B the t-value is t_o-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)

Intersection of circle and line.

 
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

Intersection points of two circles

 
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
<=>
        ...

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 |

 
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

 

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

and point P(xo,yo) is on C.

                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

 
        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

Analytic expression of the power of a point

 
        (x - a)2  + (y - b)2 - r2  = 0

 
        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

 
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

 
C1:     (x - a)2  + (y - b)2 - r2  = 0
C2:     (x - c)2  + (y - d)2 - r'2 = 0

 
  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

 
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

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.

• 
  

  A circle C has an equation (x-3)2 +(y-7)2 = 49 Find the tangent lines through point A(15,5) to C

  
  

  1. First method:

     We start with a line through A with a variable slope m.
     y - 5 = m (x - 15)
     The line is a tangent to C if and only if the two intersection points coincide.
     These intersection points are the solutions of the system

      
       / y - 5 = m (x - 15)
       |
       \ (x-3)2 +(y-7)2 = 49
     

     We substitute y from the first equation into the second equation. After working out we find

     (1 + m²) x² - (4 m + 30 m² + 6) x + 225 m² + 60 m - 36
     

     The two intersection points coincide if and only if the discriminant is zero.

     -380 m² - 192 m + 180 = 0

     The two m-values are -0.986 and 0.481
     The tangent lines are
     y - 5 = -0.986 (x - 15) and y - 5 = 0.481 (x - 15)
     

  2. Second method:

     The center of the circle is M(3,7).
     We start with a line through A with a variable slope m.
     y - 5 = m (x - 15)
     The line is a tangent to C if and only if the distance from M to the line is equal to the radius of the circle .

      
          | m 3 - 7 + 5 -15 m |
          --------------------- = 7
              sqrt( m2 + 1)
     
     <=>
     
         (12 m + 2)2 = 49 (m2 + 1)
     
     

     This is an quadratic equation in m. We find exactly the same m-values as above.

• 
  

  A variable circle c has equation x2 + y2 - 2 (t2 - 3 t + 1) x - 2 (t2 + 2 t) y + t = 0 The number t is a parameter. Calculate the point P with a constant power with respect to c. How much is that power.

  
  

  Say P has coordinates (r,s), then the power of P with respect to c is

   
       r2 + s2 - 2 (t2 - 3 t + 1) r - 2 (t2 + 2 t) s + t
  
  <=>  -(2 s + 2 r) t2 + (6 r - 4 s + 1) t + r2 + s2 - 2 r
  

  This power is independent of the parameter t if and only if

   
        2 s + 2 r = 0  and  6 r - 4 s + 1= 0
  
  <=>    r = 1/10 and s = -1/10
  

  The point P(0.1; -0.1) has a constant power with respect to the variable circle. This power is 0.22 .

• 
  

  The red astroid has parametric equations [ x = cos3(t) , y = sin3(t) ]. Find the equation of the blue circle.

  
  

  If t varies point P(cos³(t), sin³(t)) describes the astroid.
  For t = 0 we find P(1,0)
  For t = pi/2 we find P(0,1)
  For t = pi/4 we find P(2^-3/2,2^-3/2)
  So, the radius of the circle is 1/2.
  The equation of the circle is x² + y² = 1/4

• 
  

  Find the intersection points of the parabola y = x2 and the circle with parametric equations [ x = 4 cos(t) , y = 4 + 4 sin(t) ]

  
  

  The circle has cartesian equation x² + (y-4)² = 16.
  We have to find the solutions of the system

   
     x2 + (y-4)2 = 16
     y = x2
  

  The intersection points are (0,0) (sqrt(7),7) and (-sqrt(7),7)

• 
  

  Find the equation of the two circles with radius 5 and tangent to the parabola y=x2 in point T(1,1).

  
  

  In point T, the slope of the tangent line to the parabola is two. The slope of the normal line is -1/2.
  The equation of the normal line through T is y = (-1/2) x + 3/2.
  P(t, (-1/2) t + 3/2 ) is a variable point on this normal line. t is a parameter.
  P is the center of a circle if and only if

   
       |PT|2 = 25
  <=>
      (t-1)2 + ((-1/2) t + 3/2 )2 = 25
  <=>
      t = 1 ± 2.sqrt(5)
  
  Then the centers are :
  
    P1 ( 1 + 2 sqrt(5) , 1 - sqrt(5) )
    P2 ( 1 - 2 sqrt(5) , 1 + sqrt(5) )
  

  The circles have P₁ and P₂ as center and 5 as radius.

  Note the value of using a parameter.

• 
  

  The circle C1 has an equation x2 + y2 = 1 The circle C2 has an equation (x-5)2 + y2 = 9 Find the equations of the common outer tangent lines

  
  

   
         
  
  
  
     |OA| = 5 ; |AB| = 2 ; |OB| = sqrt(21)
  
     tan(t) = 2/sqrt(21) = slope  OB = slope of tangent line
  
     The equation of  OB is
  
     y = ( 2/sqrt(21) ) x
  
  <=>
     2 x - sqrt(21) y = 0
  
     The equation of the common tangent line has the form
  
     2 x - sqrt(21) y + k = 0 with k > 0
  
     We calculate the k-value such that the distance from O to the tangent line is 1
  
          k
     ---------------- = 1   <=> k = 5
     sqrt( 4 + 21 )
  
     The tangent line is  2 x - sqrt(21) y + 5 = 0
  

  Now, find the other outer tangent line

• 
  

  The circle C1 has center M1(1,1), is tangent to the x-axis and to the y-axis. Find the equation of the circle C2 such that radius > 1 center M2 on the line y=x tangent to the x-axis tangent to the y-axis tangent to C1 in punt T

  
  

  Make a clear drawing of the position of the circles.
  Let r = the radius of C₂

   
  
     M2(r,r)  =>  |O M2| = r sqrt(2)       (*)
  
  
    |O M1| = sqrt(2)
    |M1 T| = 1
    |T M2| = r
    Addition gives  sqrt(2) + 1 + r = |O M2|   (**)
  
    from (*) and (**) it follows that
  
         r sqrt(2) =  sqrt(2) + 1 + r
  <=>
         r (sqrt(2)-1) =  sqrt(2) + 1
  <=>
         r = 3 + 2 sqrt(2)
  
    and the equation is
  
    (x-r)2 +(y-r)2 = r2
  

• 
  

  Find the equation of the circle inscribed in the triangle ABC such that : AB is y = 0 AC is y = sqrt(3) x BC is y = - sqrt(3) x + 10

  
  

  Make an accurate figure.

  We'll see that the solution is easy when there is a good cooperation between geometry and analysis.

  The triangle is isosceles with top angle C and A = A(0,0) and B = B(10/sqrt(3) , 0).
  The center of the inscribed circle is on the line x = 5/sqrt(3).
  The angle A is 60 degrees.
  The bisector through A makes with the x-axis an angle of 30 degrees. This bisector has equation y = (1/sqrt(3)) x.
  

  The center M of the circle is M( 5/sqrt(3) , 5/3 ).
  The radius is 5/3.
  The equation is (x - 5/sqrt(3))² + (y - 5/3)² = 25/9

• 
  

  Take the triangle ABC with A(1,1) B(2,5) and C(7,-1). Calculate the equation of the circumscribed circle.

  
  

  From the theory it follows immediately that the equation of the circumscribed circle is

   
   | x2 + y2    x    y   1 |
   |     2      1    1   1 |
   |    29      2    5   1 |   = 0
   |    50      7   -1   1 |
  
  <=>
   13 x2 + 13 y2 - 123 x - 57 y +154 = 0
  

• 
  

  Five tangent circles with equal radius are in a square with side = 10 (see figure). Find the radius of the circles.

  
  

  Let r = the radius of the circles.
  The diagonal of the small squares is r.sqrt(2)
  The diagonal of the big square is 10.sqrt(2)
  Now, 10.sqrt(2) = 2.r.sqrt(2) + 4.r
  So, r = 5.sqrt(2)/(2 + sqrt(2))

• 
  

  Three tangent circles have radius = 5. There is a elastic band stretched around the circles (see figure). Find the length of the elastic band.

  
  

  The blue segments have length 10.
  Together, the three red arcs form exactly one circle with length 20.pi.
  The total length of the elastic band is 10 + 20.pi.

  Note the usefulness of using a geometric analysis

• 
  

  In point B(0,1), the line d with equation 2x+ y = 1 is tangent to a circle and point A(5,2) is on that circle. Find the equation of the circle.

  
  

  Draw a figure. The requested circle must satisfy two conditions:

  1. The line d is tangent in point B
  2. Point A is on the circle

  We start with the determination of all circles that meet the first condition. To this end we look for a geometric feature that allows us to find the center and the radius in a simple manner.

  The center M of the circle is on the perpendicular l through B to d.
  This perpendicular is the line y = (1/2)x + 1.
  We let M slide on the line l.
  M(2t, t+ 1) is a variable point on l. t is a parameter.

  The radius of the circle must be |M,B|.
  Now, |M,B|² = 4t² + t² = 5t².
  The equation of the circle is of the form

   
     (x - 2 t)2  + (y - t - 1)2 = 5 t2       (*)
  

  The first condition is met.

  Now we determine the value of t so that the second condition is met.

   
    A(5,2) ligt op de cirkel
  <=>
    (5 - 2 t)2  + (2 - t - 1)2 = 5 t2
  <=>
    ...
  <=>
    t = 13/11
  

  We bring this value in (*) and we have the requested circle.

  Note the usefulness of using a parameter