Solution - Properties of circles

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

$$d=2\cdot r$$

r=36.878

$$d=2\cdot 36.878$$

$$d=73.756$$

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=36.878

$$c=2\cdot 36.878\cdot \pi$$

$$c=73.756\cdot \pi$$

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

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

r=36.878

$$a={36.878}^2\cdot \pi$$

$$a=1360\cdot \pi$$

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:
$$x^2+y^2=1360$$
$$h=0$$
$$k=0$$
Center $$\left(0,0\right)$$

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 $$x$$:

$${\left(x+0\right)}^2+{\left(y+0\right)}^2=1360$$

$${\left(x+0\right)}^2+{\left(0+0\right)}^2=1360$$

$${\left(x+0\right)}^2+{\left(-0\right)}^2=1360$$

$${\left(x+0\right)}^2+0=1360$$

$${\left(x+0\right)}^2=1360-0$$

$${\left(x+0\right)}^2=1360$$

$$\sqrt{\left({\left(x+0\right)}^2\right)}=\sqrt{\left(1360\right)}$$

$$x+0=\sqrt{\left(1360\right)}$$

$$x=\pm \sqrt{\left(1360\right)}-0$$

$$x_1=\left(\sqrt{\left(1360\right)}-0,0\right),x_2=\left(-\sqrt{\left(1360\right)}-0,0\right)$$

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 $$y$$:

$${\left(x+0\right)}^2+{\left(y+0\right)}^2=1360$$

$${\left(0+0\right)}^2+{\left(y+0\right)}^2=1360$$

$${\left(-0\right)}^2+{\left(y+0\right)}^2=1360$$

$$0+{\left(y+0\right)}^2=1360$$

$${\left(y+0\right)}^2=1360-0$$

$${\left(y+0\right)}^2=1360$$

$$\sqrt{\left({\left(y+0\right)}^2\right)}=\sqrt{\left(1360\right)}$$

$$y+0=\sqrt{\left(1360\right)}$$

$$y=\pm \sqrt{\left(1360\right)}-0$$

$$y_1=\left(0,\sqrt{\left(1360\right)}-0\right),y_2=\left(0,-\sqrt{\left(1360\right)}-0\right)$$