The equation of a circle can be found using the centre and radius. The discriminant can determine the nature of intersections between two circles or a circle and a line to prove for tangency.

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.

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 \(\sqrt {36} = 6\)

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.

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 \(\sqrt {\frac{{145}}{{36}}} = \frac{{\sqrt {145} }}{6}\)

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.