Solution - Properties of circles

The diameter $$\left(d\right)$$ is equal to twice the radius:
$$d=2·r$$

$$d=2\cdot r$$

r=31.623

$$d=2\cdot 31.623$$

$$d=63.246$$

The circumference $$\left(c\right)$$ is equal to two times the radius times π:
$$c=2·r·\pi$$

$$c=2\cdot r\cdot \pi$$

r=31.623

$$c=2\cdot 31.623\cdot \pi$$

$$c=63.246\cdot \pi$$

The area $$\left(a\right)$$ is equal to the radius squared times π:
$$a=r^2·\pi$$

$$a=r^2\cdot \pi$$

r=31.623

$$a={31.623}^2\cdot \pi$$

$$a=1000\cdot \pi$$

The coordinates of the center of a circle are usually, but not always, represented by $$h$$ and $$k$$ in a circle's standard form equation:
$${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$$
Identify the $$h$$ and $$k$$ in the equation:
$${\left(y-2\right)}^2+{\left(z-22\right)}^2=1000$$
$$h=2$$
$$k=22$$
Center $$\left(2,22\right)$$

To find the $$x$$ -intercept(s), substitute $$0$$ for $$y$$ in the circle's standard form equation
$${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$$
and solve the quadratic equation for $$x$$:

$${\left(y-2\right)}^2+{\left(z-22\right)}^2=1000$$

$${\left(y-2\right)}^2+{\left(0-22\right)}^2=1000$$

$${\left(y-2\right)}^2+{\left(-22\right)}^2=1000$$

$${\left(y-2\right)}^2+484=1000$$

$${\left(y-2\right)}^2=1000-484$$

$${\left(y-2\right)}^2=516$$

$$\sqrt{\left({\left(y-2\right)}^2\right)}=\sqrt{\left(516\right)}$$

$$y-2=\sqrt{\left(516\right)}$$

$$y=\pm \sqrt{\left(516\right)}+2$$

$$y_1=\left(\sqrt{\left(516\right)}+2,0\right),y_2=\left(-\sqrt{\left(516\right)}+2,0\right)$$

To find the $$y$$ -intercept(s), substitute $$0$$ for $$x$$ in the circle's standard form equation
$${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$$
and solve the quadratic equation for $$y$$:

$${\left(y-2\right)}^2+{\left(z-22\right)}^2=1000$$

$${\left(0-2\right)}^2+{\left(z-22\right)}^2=1000$$

$${\left(-2\right)}^2+{\left(z-22\right)}^2=1000$$

$$4+{\left(z-22\right)}^2=1000$$

$${\left(z-22\right)}^2=1000-4$$

$${\left(z-22\right)}^2=996$$

$$\sqrt{\left({\left(z-22\right)}^2\right)}=\sqrt{\left(996\right)}$$

$$z-22=\sqrt{\left(996\right)}$$

$$z=\pm \sqrt{\left(996\right)}+22$$

$$z_1=\left(0,\sqrt{\left(996\right)}+22\right),z_2=\left(0,-\sqrt{\left(996\right)}+22\right)$$