Finding the Equation of a Circle from Chords
The circumcentre can be used to find the equation of a circle.
- Find three points on the circle.
- Draw chords connecting these points.
- Find the midpoints of each chord.
- Find the equations of the perpendicular bisectors.
- Find the point of intersection of these bisectors.
- The intersection point is the centre of the circle. Use this in the circle equation.
Example
Find the equation of the circle which passes through the points A(-6, -4), B(-1, -5), C(-1, 1).
Sketch the circle and draw in the chords:
Using the midpoint formula:
\[
M_{AB}\left( \frac{x_1 + x_2}{2},\; \frac{y_1 + y_2}{2} \right)
\]
\[
M_{AB}\left( \frac{-6 + (-1)}{2},\; \frac{-4 + (-5)}{2} \right)
\]
\[
M_{AB}\left( \frac{-7}{2},\; \frac{-9}{2} \right)
\]
\[
M_{BC}\left( \frac{x_1 + x_2}{2},\; \frac{y_1 + y_2}{2} \right)
\]
\[
M_{BC}\left( \frac{(-1) + (-1)}{2},\; \frac{(-5) + 1}{2} \right)
\]
\[
M_{BC}\left( -1,\; -2 \right)
\]
\[
M_{AC}\left( \frac{x_1 + x_2}{2},\; \frac{y_1 + y_2}{2} \right)
\]
\[
M_{AC}\left( \frac{-6 + (-1)}{2},\; \frac{-4 + 1}{2} \right)
\]
\[
M_{AC}\left( \frac{-7}{2},\; \frac{-3}{2} \right)
\]
Now find the gradients of the chords:
\[
m_{AB}
= \frac{y_2 - y_1}{x_2 - x_1}
\]
\[
m_{AB}
= \frac{(-5) - (-4)}{\,(-1) - (-6)\,}
\]
\[
m_{AB}
= \frac{-1}{5}
\]
\[
\text{so } m_{\perp AB} = -\frac{1}{m_{AB}} = 5
\]
\[
m_{BC}
= \frac{y_2 - y_1}{x_2 - x_1}
\]
\[
m_{BC}
= \frac{1 - (-5)}{\,(-1) - (-1)\,}
\]
\[
m_{BC}
= \frac{6}{0}
\quad\text{(vertical line)}
\]
\[
\text{so } m_{\perp BC} = 0
\]
\[
m_{AC}
= \frac{y_2 - y_1}{x_2 - x_1}
\]
\[
m_{AC}
= \frac{1 - (-4)}{-1 - (-6)}
\]
\[
m_{AC}
= \frac{5}{5} = 1
\]
\[
\text{so } m_{\perp AC}
= -\frac{1}{m_{AC}}
= -1
\]
Use these to find the perpendicular bisectors:
\[
\text{Line } \perp AB \text{ through } M_{AB}\!\left( -\frac{7}{2},\; -\frac{9}{2} \right)
\]
\[
y - b = m(x - a)
\]
\[
y + \frac{9}{2}
= 5\!\left( x + \frac{7}{2} \right)
\]
\[
y + \frac{9}{2}
= 5x + \frac{35}{2}
\]
\[
2y + 9 = 10x + 35
\]
\[
2y = 10x + 26
\]
\[
\text{Line } \perp BC \text{ through } M_{BC}(-1,\,-2)
\]
\[
y - b = m(x - a)
\]
\[
y + 2 = 0
\]
\[
y = -2
\]
\[
\text{Line } \perp AC \text{ through }
M_{AC}\!\left( -\frac{7}{2},\; -\frac{3}{2} \right)
\]
\[
y - b = m(x - a)
\]
\[
y + \frac{3}{2}
= -1\!\left( x + \frac{7}{2} \right)
\]
\[
y + \frac{3}{2}
= -x - \frac{7}{2}
\]
\[
2y + 3 = -2x - 7
\]
\[
2y = -2x - 10
\]
\[
y = -x - 5
\]
Now solve the equations to find the centre:
\[
2y = 10x + 26 \qquad (1)
\]
\[
y = -2 \qquad (2)
\]
\[
2y = -2x - 10 \qquad (3)
\]
\[
\text{Equate (1) and (3)}
\]
\[
10x + 26 = -2x - 10
\]
\[
12x = -36
\]
\[
x = -3
\]
\[
\text{Centre is } (-3,\,-2)
\]
Next, find the radius using the distance formula.
\[
AD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
\[
AD = \sqrt{\bigl(-3 - (-6)\bigr)^2 + \bigl(-2 - (-4)\bigr)^2}
\]
\[
AD = \sqrt{3^2 + 2^2}
\]
\[
AD = \sqrt{13}
\]
\[
\text{radius} = \sqrt{13}
\]
Finally, substitute the centre and radius into the equation of a circle:
\[
\text{Centre } (-3,\,-2),
\qquad
\text{radius } = \sqrt{13}
\]
\[
(x - a)^2 + (y - b)^2 = r^2
\]
\[
(x + 3)^2 + (y + 2)^2 = 13
\]
