# How do you find the center and radius for x^2 - 3x + y^2 - 3y - 20 = 0?

Jul 28, 2018

#### Answer:

The center is $= \left(\frac{3}{2} , \frac{3}{2}\right)$ and the radius $r = 7 \frac{\sqrt{2}}{2}$

#### Explanation:

The standard equation of a circle is

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

where

The center is $\left(a , b\right)$ and the radius is $= r$

Here, we have

${x}^{2} - 3 x + {y}^{2} - 3 y - 20 = 0$

Complete the square for $x$ and $y$

${x}^{2} - 3 x + \frac{9}{4} + {y}^{2} - 3 y + \frac{9}{4} = 20 + \frac{9}{4} + \frac{9}{4}$

${\left(x - \frac{3}{2}\right)}^{2} + {\left(y - \frac{3}{2}\right)}^{2} = \frac{98}{4} = {\left(7 \frac{\sqrt{2}}{2}\right)}^{2}$

The center is $= \left(\frac{3}{2} , \frac{3}{2}\right)$ and the radius $r = 7 \frac{\sqrt{2}}{2}$

graph{x^2-3x+y^2-3y-20=0 [-16.35, 15.68, -4.36, 11.66]}