Equation of a Circle through Three Points

Consider the general equation 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 circle. Now put the above three points in the given equation of 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 required circle, we must the find values of g,f,c from the above equation (i), (ii) and (iii) and put these values back in the general equation of circle, for this using the method of solving simultaneous equations also we can use methods of matrix like Cramer’s Rule.

Example: Find the equation of 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 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) lies on the circle and now put these points in the above equation of 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 solving the equations (i) and (ii), by subtracting equation (ii) and (i) we get the new equation as

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


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

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


Now solving the equation (iv) and (v), and we the values of g and f as f  =  - \frac{{11}}{6} and g =  -  \frac{{13}}{6} now putting these calculated values in equation (i) so have the value of c = \frac{{20}}{3}.
Now putting all these three values in the first equation (A), to get the required equation of 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}

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