Solution - Properties of circles

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

$$d=2\cdot r$$

r=11.136

$$d=2\cdot 11.136$$

$$d=22.271$$

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=11.136

$$c=2\cdot 11.136\cdot \pi$$

$$c=22.271\cdot \pi$$

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

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

r=11.136

$$a={11.136}^2\cdot \pi$$

$$a=124\cdot \pi$$

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:
$${\left(x+1\right)}^2+{\left(y-2\right)}^2=124$$
$$h=-1$$
$$k=2$$
Center $$\left(-1,2\right)$$

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 $$x$$:

$${\left(x+1\right)}^2+{\left(y-2\right)}^2=124$$

$${\left(x+1\right)}^2+{\left(0-2\right)}^2=124$$

$${\left(x+1\right)}^2+{\left(-2\right)}^2=124$$

$${\left(x+1\right)}^2+4=124$$

$${\left(x+1\right)}^2=124-4$$

$${\left(x+1\right)}^2=120$$

$$\sqrt{\left({\left(x+1\right)}^2\right)}=\sqrt{\left(120\right)}$$

$$x+1=\sqrt{\left(120\right)}$$

$$x=\pm \sqrt{\left(120\right)}-1$$

$$x_1=\left(\sqrt{\left(120\right)}-1,0\right),x_2=\left(-\sqrt{\left(120\right)}-1,0\right)$$

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 $$y$$:

$${\left(x+1\right)}^2+{\left(y-2\right)}^2=124$$

$${\left(0+1\right)}^2+{\left(y-2\right)}^2=124$$

$${\left(1\right)}^2+{\left(y-2\right)}^2=124$$

$$1+{\left(y-2\right)}^2=124$$

$${\left(y-2\right)}^2=124-1$$

$${\left(y-2\right)}^2=123$$

$$\sqrt{\left({\left(y-2\right)}^2\right)}=\sqrt{\left(123\right)}$$

$$y-2=\sqrt{\left(123\right)}$$

$$y=\pm \sqrt{\left(123\right)}+2$$

$$y_1=\left(0,\sqrt{\left(123\right)}+2\right),y_2=\left(0,-\sqrt{\left(123\right)}+2\right)$$