Solution - Properties of circles

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

$$d=2\cdot r$$

r=17.088

$$d=2\cdot 17.088$$

$$d=34.176$$

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=17.088

$$c=2\cdot 17.088\cdot \pi$$

$$c=34.176\cdot \pi$$

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

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

r=17.088

$$a={17.088}^2\cdot \pi$$

$$a=292\cdot \pi$$

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:
$${\left(x+1\right)}^2+{\left(y+1\right)}^2=292$$
$$h=-1$$
$$k=-1$$
Center $$\left(-1,-1\right)$$

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 $$x$$:

$${\left(x+1\right)}^2+{\left(y+1\right)}^2=292$$

$${\left(x+1\right)}^2+{\left(0+1\right)}^2=292$$

$${\left(x+1\right)}^2+{\left(1\right)}^2=292$$

$${\left(x+1\right)}^2+1=292$$

$${\left(x+1\right)}^2=292-1$$

$${\left(x+1\right)}^2=291$$

$$\sqrt{\left({\left(x+1\right)}^2\right)}=\sqrt{\left(291\right)}$$

$$x+1=\sqrt{\left(291\right)}$$

$$x=\pm \sqrt{\left(291\right)}-1$$

$$x_1=\left(\sqrt{\left(291\right)}-1,0\right),x_2=\left(-\sqrt{\left(291\right)}-1,0\right)$$

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 $$y$$:

$${\left(x+1\right)}^2+{\left(y+1\right)}^2=292$$

$${\left(0+1\right)}^2+{\left(y+1\right)}^2=292$$

$${\left(1\right)}^2+{\left(y+1\right)}^2=292$$

$$1+{\left(y+1\right)}^2=292$$

$${\left(y+1\right)}^2=292-1$$

$${\left(y+1\right)}^2=291$$

$$\sqrt{\left({\left(y+1\right)}^2\right)}=\sqrt{\left(291\right)}$$

$$y+1=\sqrt{\left(291\right)}$$

$$y=\pm \sqrt{\left(291\right)}-1$$

$$y_1=\left(0,\sqrt{\left(291\right)}-1\right),y_2=\left(0,-\sqrt{\left(291\right)}-1\right)$$