# Equation of a circle

The general equation of a circle normally appears in the form $${x^2} + {y^2} + 2gx + 2fy + c = 0$$

where $$( - g, - f)$$ is the centre of the circle

and $$\sqrt {{g^2} + {f^2} - c}$$ is the radius.

Notice that for the circle to exist, $${g^2} + {f^2} - c\textgreater0$$.

Look at the following worked examples.

## Example 1

For $${x^2} + {y^2} + 6x - 8y - 11 = 0$$

${g^2} + {f^2} - c = {(3)^2} + {( - 4)^2} - ( - 11) = 36$

So the equation represents a circle with centre $$( - 3,4)$$ and radius $$\sqrt {36} = 6$$

## Example 2

For $${x^2} + {y^2} - 2x + 4y + 11 = 0$$

${g^2} + {f^2} - c = {( - 1)^2} + {(2)^2} - 11 = - 6$

So $${x^2} + {y^2} - 2x + 4y + 11 = 0$$ does not represent a circle.

## Example 3

For $$3{x^2} + 3{y^2} - 6x + y - 9 = 0$$ we must write this starting with $${x^2} + {y^2}$$:

${x^2} + {y^2} - 2x + \frac{1}{3}y - 3 = 0$

${g^2} + {f^2} - c = {( - 1)^2} + {\left( {\frac{1}{6}} \right)^2} - ( - 3)$

$= 4\frac{1}{{36}} = \frac{{145}}{{36}}$

So the equation represents a circle with centre $$\left( {1, - \frac{1}{6}} \right)$$ and radius $$\sqrt {\frac{{145}}{{36}}} = \frac{{\sqrt {145} }}{6}$$

## Questions

For each of the following equations, state whether it could represent a circle and if so, state the radius and centre.

Question

$2{x^2} + 2{y^2} + 4x - 3y - 6 = 0$

${x^2} + {y^2} + 2x - \frac{3}{2}y - 3 = 0$

${g^2} + {f^2} - c = {1^2} + {\left( { - \frac{3}{4}} \right)^2} - ( - 3) = \frac{{73}}{{16}}$

So the equation represents a circle with centre $$\left( { - 1,\frac{3}{4}} \right)$$ and radius $$\frac{{\sqrt {73} }}{4}$$

Question

${x^2} + {y^2} + 2x - 4y + 6 = 0$

${g^2} + {f^2} - c = {(1)^2} + {\left( { - 2} \right)^2} - 6 = - 1$

So the equation does not represent a circle.