# General Form of the Equation of a Circle

Consider the equation of a circle in general form is

\[\boxed{{x^2} + {y^2} + 2gx + 2fy + c = 0}\,\,\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)\]

Where $$g,f,c$$ are any constant values.

If we rearrange the terms of the above equation (i) of a circle, we have

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

In this equation we use the method of completing squares, so for this we need to add $${g^2}$$ and $${f^2}$$ on both sides of the equation (ii). i.e.:

\[\begin{gathered} {x^2} + 2gx + {g^2} + {y^2} + 2fy + {f^2} + c = {g^2} + {f^2} \\ \Rightarrow {\left( {x + g} \right)^2} + {\left( {y + f} \right)^2} = {g^2} + {f^2} – c \\ \Rightarrow {\left[ {x – \left( { – g} \right)} \right]^2} + {\left[ {y – \left( { – f} \right)} \right]^2} = {\left[ {\sqrt {{g^2} + {f^2} – c} } \right]^2} \\ \end{gathered} \]

Compare this equation of a circle with the standard equation of a circle $${\left( {x – h} \right)^2} + {\left( {y – k} \right)^2} = {r^2}$$ and we get the radius $$\sqrt {{g^2} + {f^2} – c} $$ and centre $$\left( { – g, – f} \right)$$.

This shows that the equation $${x^2} + {y^2} + 2gx + 2fy + c = 0$$ represents a circle with centre $$\left( { – g, – f} \right)$$ and radius $$\sqrt {{g^2} + {f^2} – c} $$. This is called the general equation of a circle.

__Example__**:** Find the centre and radius of the circle with the given equation of a circle

\[7{x^2} + 7{y^2} + 18x – 10y + 14 = 0\]

__Solution__**:** The given equation of a circle is

\[7{x^2} + 7{y^2} + 18x – 10y + 14 = 0\]

We observe that in this equation of a circle the coefficients of $${x^2}$$ and $${y^2}$$ is **7**, but in the general form of the equation of a circle the coefficients must be equal to **1**.

To convert the given equation into the form of general equation, divide the given equation on both sides by **7.** We get

\[{x^2} + {y^2} + \frac{{18}}{7}x – \frac{{10}}{7}y + 2 = 0\]

The above equation can be written as

\[{x^2} + {y^2} + 2\left( {\frac{9}{7}} \right)x + 2\left( { – \frac{5}{7}} \right)y + 2 = 0\]

Compare this equation with the general equation of a circle as

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

We have the values $$g = \frac{9}{7}$$, $$f = – \frac{5}{7}$$ and $$c = 2$$.

Hence the centre of the circle is $$\left( { – g, – f} \right) = \left( { – \frac{9}{7}, – \left( { – \frac{5}{7}} \right)} \right) = \left( { – \frac{9}{7},\frac{5}{7}} \right)$$

The radius of the circle is $$r = \sqrt {{g^2} + {f^2} – c} = \sqrt {{{\left( { – \frac{9}{7}} \right)}^2} + {{\left( {\frac{5}{7}} \right)}^2} – 2} = \frac{{2\sqrt 2 }}{7}$$