# 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.

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

hencel

January 28@ 1:35 pmhow about (1,8),(-6,1),(-2,1)