Equation of Circle/Cartesian/Formulation 1

Theorem

The equation of a circle embedded in the Cartesian plane with radius $R$ and center $\tuple {a, b}$ can be expressed as:

$\paren {x - a}^2 + \paren {y - b}^2 = R^2$

Proof

Let the point $\tuple {x, y}$ satisfy the equation:

$(1): \quad \paren {x - a}^2 + \paren {y - b}^2 = R^2$

By the Distance Formula, the distance between this $\tuple {x, y}$ and $\tuple {a, b}$ is:

$\sqrt {\paren {x - a}^2 + \paren {y - b}^2}$

But from equation $(1)$, this quantity equals $R$.

Therefore the distance between points satisfying the equation and the center is constant and equal to the radius.

Thus $\tuple {x, y}$ lies on the circumference of a circle with radius $R$ and center $\tuple {a, b}$.

Now suppose that $\tuple {x, y}$ does not satisfy the equation:

$\paren {x - a}^2 + \paren {y - b}^2 = R^2$

Then by the same reasoning as above, the distance between $\tuple {x, y}$ and $\tuple {a, b}$ does not equal $R$.

Therefore $\tuple {x, y}$ does not lie on the circumference of a circle with radius $R$ and center $\tuple {a, b}$.

Hence it follows that the points satisfying $(1)$ are exactly those points which are the circle in question.

$\blacksquare$