The Circle -- Basic concepts




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).

Special Case :

Each circle with center O(0,0) has an equation of the form

 
         x2 + y2 = r2
Example :

We calculate the equation of the circle with center A(1,2) and through point B(4,6).

The radius of the circle is |A,B| = sqrt((4-1)2 + (6-2)2 ) = 5.

The equation is (x-1)2 + (y-2)2 = 25

This equation is equivalent with x2 +y2 - 2x - 4y - 20 = 0

Inverse property

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

Examples

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 :

A circle C with center M(3,-2) and radius 4 has parametric equations

 
   x = 3 + 4 cos(t)
   y = -2 + 4 sin(t)
Each t-value corresponds with a point of the circle.

Exercise: Draw the circle C. Choose a few t-values and calculate the coordinates of the corresponding point P. Check if point P is on the circle C.

Example 2 :

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)

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).

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) 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
Example :

Let C be the circle with center M(3,4) and radius r = 5. Find the equation of the tangent line in point O(0,0).

Solution:

The equation of the circle is (x-3)2 + (y-4)2 = 25

The tangent line in point O(0,0) is

 
                  3 - 0
      y - 0 = -  ------- (x - 0)
                  4 - 0

<=>    y = (-3/4) x

<=>    3x + 4 y = 0

Length of a circular arc

The circumference of a circle with radius r is 2.pi.r
Consider a circular arc corresponing with a central angle of t radians. The length of the arc is directly proportional to the angle t. The length is
 
                  t
     (2 pi r) . ------  = r t
                 2 pi
A circle has radius r.
The length of the arc, corresponing with a central angle of t radians, is r.t

Area of a circle sector

The area of a circle with radius r is pi.r2
Consider a circle sector with a central angle of t radians. The area of the sector is directly proportional to the angle t. The area is
 
                  t
     (pi r2) . ------  = (1/2) t r2
                 2 pi
A circle has radius r.
The area of the sector with a central angle of t radians is (1/2) t r2

More about the circle

This introduction page showed how basic properties of a circle can be expressed in an analytic way.

More complex analytic geometry about the circle can be found on a second page about the circle.

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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