Radius of circle from quadratic equation

The equation of a circle of radius $r$ and centre the origin is $x^2 + y^2 = r^2$ . The given equation in the problem is $x^2 + y^2 = 9$ . We see that $r^2 = 9$ , so the radius is,

$r = \sqrt{9} = \boxed{3}$

The simplest way to solve the problem would be to find the x or y-intercept of the equation.

Y-intercepts: ( $x=0$ )

$(0)^2 + y^2 = 9$

$y=±3$

The circle passes through $(0, 3)$ and $(0, -3)$ , and the distance between these points is the circle's diameter.

$3-(-3) = 6$ (diameter)

${6\over 2} = 3$ (radius, which is 0.5 * diameter)