Equation of a Circle through Three Points

Consider the general equation a circle is given by

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


three-points-eq-circle

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}