# Equation of a Circle through Three Points

Consider the general equation a circle is given by

${x^2} + {y^2} + 2gx + 2fy + c = 0$

If the given circle is passing through three non-collinear points, say, $A\left( {{x_1},{y_1}} \right)$, $B\left( {{x_2},{y_2}} \right)$ and $C\left( {{x_3},{y_3}} \right)$, then these points must satisfy the general equation of a circle. Now put the above three points in the given equation of a circle, i.e.:

$\begin{gathered} {x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + c = 0\,\,\,{\text{ – – – }}\,\left( {\text{i}} \right) \\ {x_2}^2 + {y_2}^2 + 2g{x_2} + 2f{y_2} + c = 0\,\,\,{\text{ – – – }}\,\left( {{\text{ii}}} \right) \\ {x_3}^2 + {y_3}^2 + 2g{x_3} + 2f{y_3} + c = 0\,\,\,{\text{ – – – }}\,\left( {{\text{iii}}} \right) \\ \end{gathered}$

To evaluate the equation of the required circle, we must the find the values of $g,f,c$ from the above equations (i), (ii) and (iii), and put these values back in the general equation of a circle. Using this method of solving simultaneous equations we can also use methods of a matrix like Cramer’s Rule.

Example: Find the equation of a circle through three non-collinear points $\left( {1,2} \right)$, $\left( {2,3} \right)$ and $\left( {3,1} \right)$.

Solution: Consider the required equation of a circle in general form as

${x^2} + {y^2} + 2gx + 2fy + c = 0\,\,\,{\text{ – – – }}\left( {\text{A}} \right)$

Since the given points $\left( {1,2} \right)$, $\left( {2,3} \right)$ and $\left( {3,1} \right)$ lie on the circle, putting these points in the above equation of a circle (A) becomes for these three points:

$\begin{gathered} 5 + 2g + 4f + c = 0\,\,\,{\text{ – – – }}\left( {\text{i}} \right) \\ 13 + 4g + 6f + c = 0\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right) \\ 10 + 6g + 2f + c = 0\,\,\,{\text{ – – – }}\left( {{\text{iii}}} \right) \\ \end{gathered}$

First, by solving equations (i) and (ii) and by subtracting equation (ii) and (i) we get the new equation as

$8 + 2g + 2f = 0\,\,\,{\text{ – – – }}\left( {{\text{iv}}} \right)$

Also by solving equations (ii) and (iii) and by subtracting equation (ii) and (iii) we get the new equation as

$3 – 2g + 4f = 0\,\,\,{\text{ – – – }}\left( {\text{v}} \right)$

Now we solve equations (iv) and (v), and we the values of $g$ and $f$ as $f = – \frac{{11}}{6}$ and $g = – \frac{{13}}{6}$. We put these calculated values in equation (i) so we have the value of $c = \frac{{20}}{3}$.

Now we put all these three values in the first equation (A) to get the required equation of a circle passing through three non-collinear points.

$\begin{gathered} {x^2} + {y^2} + 2\left( { – \frac{{13}}{6}} \right)x + 2\left( { – \frac{{11}}{6}} \right)y + \frac{{20}}{3} = 0 \\ 3{x^2} + 3{y^2} – 13x – 11y + 20 = 0 \\ \end{gathered}$