Equation of a circle
Watch this video to learn about the 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.
