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.