Solution - Properties of circles

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

$$d=2\cdot r$$

r=2

$$d=2\cdot 2$$

$$d=4$$

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=2

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

$$c=4\cdot \pi$$

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

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

r=2

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

$$a=4\cdot \pi$$

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:
$$x^2+{\left(y-2\right)}^2=4$$
$$h=0$$
$$k=2$$
Center $$\left(0,2\right)$$

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 $$x$$:

$${\left(x+0\right)}^2+{\left(y-2\right)}^2=4$$

$${\left(x+0\right)}^2+{\left(0-2\right)}^2=4$$

$${\left(x+0\right)}^2+{\left(-2\right)}^2=4$$

$${\left(x+0\right)}^2+4=4$$

$${\left(x+0\right)}^2=4-4$$

$${\left(x+0\right)}^2=0$$

$$\sqrt{\left({\left(x+0\right)}^2\right)}=\sqrt{\left(0\right)}$$

$$x+0=\sqrt{\left(0\right)}$$

$$x=\pm \sqrt{\left(0\right)}-0$$

$$x=\pm 0-0$$

$$x=\left(0,0\right)$$

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 $$y$$:

$${\left(x+0\right)}^2+{\left(y-2\right)}^2=4$$

$${\left(0+0\right)}^2+{\left(y-2\right)}^2=4$$

$${\left(-0\right)}^2+{\left(y-2\right)}^2=4$$

$$0+{\left(y-2\right)}^2=4$$

$${\left(y-2\right)}^2=4-0$$

$${\left(y-2\right)}^2=4$$

$$\sqrt{\left({\left(y-2\right)}^2\right)}=\sqrt{\left(4\right)}$$

$$y-2=\sqrt{\left(4\right)}$$

$$y=\pm \sqrt{\left(4\right)}+2$$

$$y=\pm 2+2$$

$$y_1=\left(0,0\right),y_2=\left(0,4\right)$$