What Is the Equation of a Circle?

Circle with radius r

Rule

The Circle

The equation of a circle with its center at $(x_{0}, y_{0})$ and with radius $r$ is given by

{(x - x_{0})}^{2} + {(y - y_{0})}^{2} = r^{2}

This means that all values for $(x, y)$ that fulfill the equation are on the circle boundary.

To find the radius of the circle you can use this formula:

r = \sqrt{{(x - x_{0})}^{2} + {(y - y_{0})}^{2}}

Example 1

You are working with a circle with its center at $(- 1, 2)$ . The point $(3, 5)$ is on the circle boundary. What is the radius of this circle?

Here you have $(x_{0}, y_{0}) = (- 1, 2)$ and the point on the circle is $(x, y) = (3, 5)$ . The radius of the circle is therefore

r = \sqrt{{(3 + 1)}^{2} + {(5 - 2)}^{2}} = 5

Example 2

Show that $x^{2} + 6 x + y^{2} - 2 y - 6 = 0$ is a circle

To show this, you have to use the method to complete the square. You add and subtract ${(\frac{b}{2})}^{2}$ for both the $x$ -terms and the $y$ -terms. You then get the following:

\begin{array}{l} 0 & = x^{2} + 6 x + y^{2} - 2 y - 6, \\ = x^{2} + 6 x + {(\frac{6}{2})}^{2} - {(\frac{6}{2})}^{2} \\ + y^{2} - 2 y + {(\frac{- 2}{2})}^{2} - {(\frac{- 2}{2})}^{2} \\ - 6 \\ = x^{2} + 6 x + 9 - 9 \\ + y^{2} - 2 y + 1 - 1 - 6 \\ = {(x + 3)}^{2} + {(y - 1)}^{2} - 9 - 1 - 6 \\ = {(x + 3)}^{2} + {(y - 1)}^{2} - 16 \end{array}

\begin{array}{l} 0 & = x^{2} + 6 x + y^{2} - 2 y - 6 \\ = x^{2} + 6 x + {(\frac{6}{2})}^{2} - {(\frac{6}{2})}^{2} + y^{2} - 2 y + {(\frac{- 2}{2})}^{2} - {(\frac{- 2}{2})}^{2} - 6 \\ = x^{2} + 6 x + 9 - 9 + y^{2} - 2 y + 1 - 1 - 6 \\ = {(x + 3)}^{2} + {(y - 1)}^{2} - 9 - 1 - 6 \\ = {(x + 3)}^{2} + {(y - 1)}^{2} - 16 \end{array}

Thus,

{(x + 3)}^{2} + {(y - 1)}^{2} = 4^{2}

You can see from this that $x^{2} + 6 x + y^{2} - 2 y - 6 = 0$ is a circle with its center at $(- 3, 1)$ and a radius $r = 4$ .

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