# Equation of a Circle Through Two Points and a Line Passing Through its Center

Consider the general equation a circle is given by
${x^2} + {y^2} + 2gx + 2fy + c = 0$ If the given circle is passing through two points, say $A\left( {{x_1},{y_1}} \right)$and $B\left( {{x_2},{y_2}} \right)$, then these points must satisfy the general equation of a circle. Now put these two points in the given equation of a circle, i.e.:

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

Also, the given straight line $ax + by + {c_1} = 0$ passes through the center $\left( { – g, – f} \right)$ of the circle.

$\begin{gathered} \Rightarrow a\left( { – g} \right) + b\left( { – f} \right) + {c_1} = 0\, \\ \Rightarrow \, – ag – bf + {c_1} = 0\,\,\,{\text{ – – – }}\,\,\,\left( {{\text{iii}}} \right) \\ \end{gathered}$

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 in the first equation of a circle. We can solve these three using the method of simultaneous equations.

Example: Find the equation of a circle through two points$\left( {1,2} \right)$,$\left( {2,3} \right)$ and whose center is on the straight line $x – y + 1 = 0$.

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

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

We know that the given points $\left( {1,2} \right)$, $\left( {2,3} \right)$ lie on the circle, and if we put these points in the above equation of a circle, (A) becomes for these three points:

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

Since the center of the circle is $\left( { – g, – f} \right)$ and this center lies on the given straight line, $\left( { – g, – f} \right)$ must satisfy the equation of a line as

$– g + f + 1 = 0\,\,\,{\text{ – – – }}\left( {{\text{iii}}} \right)$

Now we solve equations (i), (ii) and (ii), using the method of simultaneous equations and we use the values of $f = – \frac{5}{2}$ ,$g = – \frac{3}{2}$ and $c = 8$

Now putting these three values in the first equation (A), we get the required equation of a circle passing through two points and with its center lying on the line.

${x^2} + {y^2} – 3x – 5y + 8 = 0$

This is the required equation of a circle.