# How do you find the center and radius of the circle x^2+5x+y^2-6y=3?

Sep 1, 2016

Center is $\left(- \frac{5}{2} , 3\right)$ and radius is $\frac{\sqrt{73}}{2}$

#### Explanation:

The equation of a circle is of the type ${x}^{2} + {y}^{2} + 2 g x + 2 f y - c = 0$. In this as the coefficients of ${x}^{2}$ and ${y}^{2}$ are equal and the term containing $x y$ is not there, equation of a circle can be reduced to above form just by dividing by coefficient if $x$ or $y$.

Now ${x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0$ can be rewritten as

${x}^{2} + 2 g x + {g}^{2} + {y}^{2} + 2 f y + {f}^{2} + c = {g}^{2} + {f}^{2}$ or

${\left(x + g\right)}^{2} + {\left(y + f\right)}^{2} = {g}^{2} + {f}^{2} - c$ or

${\left(x - \left(- g\right)\right)}^{2} + {\left(y - \left(- f\right)\right)}^{2} = {\left(\sqrt{{g}^{2} + {f}^{2} - c}\right)}^{2}$

which is equation of a circle with center at $\left(- g , - f\right)$ and radius sqrt((g^2+f^2-c).

Hence, in the equation ${x}^{2} + 5 x + {y}^{2} - 6 y = 3$ or ${x}^{2} + {y}^{2} + 5 x - 6 y - 3 = 0$, as $g = \frac{5}{2}$, $f = - 3$ and $c = - 3$,

Center is $\left(- \frac{5}{2} , 3\right)$ and radius is $\sqrt{{\left(- \frac{5}{2}\right)}^{2} + {3}^{2} - \left(- 3\right)}$

= $\sqrt{\frac{25}{4} + 9 + 3}$

= $\sqrt{\frac{73}{4}} = \frac{\sqrt{73}}{2}$

Note - One can also make complete squares in the given equation directly to find center and radius, as follows

${x}^{2} + 5 x + {y}^{2} - 6 y = 3$ or

${x}^{2} + 5 x + \frac{25}{4} + {y}^{2} - 6 y + 9 = \frac{25}{4} + 9 + 3$ or

${\left(x + \frac{5}{2}\right)}^{2} + {\left(y - 3\right)}^{2} = \frac{73}{4}$ or

${\left(x - \left(- \frac{5}{2}\right)\right)}^{2} + {\left(y - 3\right)}^{2} = {\left(\frac{\sqrt{73}}{2}\right)}^{2}$

and hence the result.