You probably remember from high school geometry that only one circle can be defined or drawn any through any three points not in a straight line.
So, if you input 3 points, this will compute the circle’s center point, radius and equation.
Example: Input these three points (5, 8) (7, 6) and (9, 2) click “ENTER”
The circle’s center is located at (-2, -1) with radius = 11.402 and
the circle’s equation is (x +2)² + (y +1)² = 130
Calculating A Circle’s Equation From 3 Points
Let’s take three points and find a circle’s center and equation.
Point A (9, 2)
Point B (3, -4)
Point C (5, -6)
Let’s put these points into ‘x’ and ‘y’ coordinates.
| x1 = 9 | y1 = 2 |
| x2 = 3 | y2 = -4 |
| x3 = 5 | y3 = -6 |
First, we will need to determine the slopes of two lines – lines AB and BC.
Slope Line AB = (y1 -y2) ÷ (x1 -x2) = (2 –4) ÷ (9 -3) = (6) ÷ (6) = 1
Slope Line BC = (y3 -y2) ÷ (x3 -x2) = (-6 –4) ÷ (5 -3) = (-2) ÷ (2) = -1
Now, we need to find the ‘x’ coordinate of the circle’s center which is:
xctr = [slope AB * slope BC * (Y3 -Y1) + slope AB * (X2 +X3) -slope BC * (X1 +X2)] ÷ [2 * (slope AB -slope BC)]
xctr = [ (1 * -1 * (-6 -2)) + (1 * (3 + 5)) – (-1 * (9 + 3))] ÷ 2 * (1 –1)
xctr = (-1 * (-8)) + 8 -(-12) ÷ 4
xctr = (8 + 8 + 12) ÷ 4
xctr = 7
To find the ‘y’ coordinate of the circle’s center we use this formula:
yctr = -(1/slope AB)*(xctr-[(x1 +x2)/2)] + (y1 +y2)/2
yctr = -(1/1) * (7 -[(9 +3)/2]) + (2 -4)/2
yctr = (-1 * (7 -6)) -(2/2)
yctr = (-1 * 1) -1
yctr = -2
Circle’s Center is located at: (7, -2)
Finally, to calculate the circle’s radius, we use this formula:
radius = Square Root [(x1 -xCtr)^2 + (y1 -yCtr)^2)]
where (x1, y1) can be any of the three points but let’s use (9, 2)
radius = Square Root [(9 -7)^2 + (2 –2)^2)]
radius = Square Root [(2)^2 + (4)^2)]
radius = Square Root (20)
radius = 4.472135955
To calculate the circle’s equation, insert those three numbers into this equation.
(x -7)² + (y –2)² = 4.472135955²
(x -7)² + (y +2)² = 20
For a tutorial about circle equations, click here.
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