# What is the center, radius, general form, and standard form of x^2 + y^2 - 2x +6y - 3 =0?

Dec 21, 2016

General form is ${\left(x - 1\right)}^{2} + {\left(y + 3\right)}^{2} = {\left(\sqrt{13}\right)}^{2}$.

This is the equation of a circle, whose center is $\left(1 , - 3\right)$ and radius is $\sqrt{13}$.

#### Explanation:

As there is no term in the quadratic equation ${x}^{2} + {y}^{2} - 2 x + 6 y - 3 = 0$ and coefficients of ${x}^{2}$ and ${y}^{2}$ are equal,

the equation represents a circle.

Let us complete the squares and see the results

${x}^{2} + {y}^{2} - 2 x + 6 y - 3 = 0$

$\Leftrightarrow {x}^{2} - 2 x + {1}^{2} + {y}^{2} + 6 y + {3}^{2} = {1}^{2} + {3}^{2} + 3 = 13$

or ${\left(x - 1\right)}^{2} + {\left(y + 3\right)}^{2} = {\left(\sqrt{13}\right)}^{2}$

It is the equation of a point which moves so that its distance from point $\left(1 , - 3\right)$ is always $\sqrt{13}$ and hence equation represents a circle, whose radius is $\sqrt{13}$.