Solution - Properties of circles

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

$$d=2\cdot r$$

r=4

$$d=2\cdot 4$$

$$d=8$$

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=4

$$c=2\cdot 4\cdot \pi$$

$$c=8\cdot \pi$$

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

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

r=4

$$a=4^2\cdot \pi$$

$$a=16\cdot \pi$$

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:
$${\left(x-6\right)}^2+{\left(y+-\right)}^2=16$$
$$h=6$$
$$k=0$$
Center $$\left(6,0\right)$$

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 $$x$$:

$${\left(x-6\right)}^2+{\left(y+0\right)}^2=16$$

$${\left(x-6\right)}^2+{\left(0+0\right)}^2=16$$

$${\left(x-6\right)}^2+{\left(-0\right)}^2=16$$

$${\left(x-6\right)}^2+0=16$$

$${\left(x-6\right)}^2=16-0$$

$${\left(x-6\right)}^2=16$$

$$\sqrt{\left({\left(x-6\right)}^2\right)}=\sqrt{\left(16\right)}$$

$$x-6=\sqrt{\left(16\right)}$$

$$x=\pm \sqrt{\left(16\right)}+6$$

$$x=\pm 4+6$$

$$x_1=\left(2,0\right),x_2=\left(10,0\right)$$

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 $$y$$:

$${\left(x-6\right)}^2+{\left(y+0\right)}^2=16$$

$${\left(0-6\right)}^2+{\left(y+0\right)}^2=16$$

$${\left(-6\right)}^2+{\left(y+0\right)}^2=16$$

$$36+{\left(y+0\right)}^2=16$$

$${\left(y+0\right)}^2=16-36$$

$${\left(y+0\right)}^2=-20$$

$$\sqrt{\left({\left(y+0\right)}^2\right)}=\sqrt{\left(-20\right)}$$

$$y+0=\sqrt{\left(-20\right)}$$

$$y=\pm \sqrt{\left(-20\right)}-0$$

No y-intercepts