# How do you find the center and radius for x^2+y^2-2x-8y+16=0?

Jun 5, 2016

C(1;4) and r=1

#### Explanation:

Center coordinates are (-a/2;-b/2) where a and b are the coefficients for x and y, respectively, in the equation;

$r = \frac{1}{2} \sqrt{{a}^{2} + {b}^{2} - 4 c}$

where c is the constant term

so

$r = \frac{1}{2} \sqrt{4 + 64 - 4 \cdot 16}$
$r = \frac{1}{2} \sqrt{4}$
$r = \frac{1}{2} \cdot 2 = 1$