# How do you find the center and radius of the circle given x^2+y^2-3x+8y=20?

Nov 24, 2016

The center is $\left(\frac{3}{2} , - 4\right)$
The radius is $= 6.185$

#### Explanation:

We have to rearrange the equation and complete the squares

${x}^{2} + {y}^{2} - 3 x + 8 y = 20$

${x}^{2} - 3 x + {y}^{2} + 8 y = 20$

${x}^{2} - 3 x + \frac{9}{4} + {y}^{2} + 8 y + 16 = 20 + \frac{9}{4} + 16$

${\left(x - \frac{3}{2}\right)}^{2} + {\left(y + 4\right)}^{2} = 38.25 = {6.185}^{2}$

So the center of the circle is $\left(\frac{3}{2} , - 4\right)$
and the radius is $= 6.185$

graph{x^2+y^2-3x+8y=20 [-14.28, 14.2, -10.85, 3.39]}