..Equation
of a circle..Theorem..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
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 circlesC1: (x - a)2 + (y - b)2 - r2 = 0 C2: (x - c)2 + (y - d)2 - r'2 = 0We 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.
Information Provided by Johan.Claeys@ping.be