Oplossing - Properties van cirkels

De diameter $$\left(d\right)$$ is equal naar twice de radius:
$$d=2·r$$

$$d=2\cdot r$$

r=9,797958971132712

$$d=2\cdot 9,797958971132712$$

$$d=19,595917942265423$$

De circumference $$\left(c\right)$$ is equal naar two times de radius times π:
$$c=2·r·\pi$$

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

r=9,797958971132712

$$c=2\cdot 9,797958971132712\cdot \pi$$

$$c=19,595917942265423\cdot \pi$$

De area $$\left(a\right)$$ is equal naar de radius squared times π:
$$a=r^2·\pi$$

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

r=9,797958971132712

$$a=9,{797958971132712}^2\cdot \pi$$

$$a=96\cdot \pi$$

De coordinates van de center van een cirkel zijn usually, but Neet always, represented door $$h$$ en $$k$$ in een cirkel's standaard form vergelijking:
$${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$$
Identify de $$h$$ en $$k$$ in de vergelijking:
$${}_{T}OK{0}_{}$$
$${}_{T}OK{1}_{}$$
$${}_{T}OK{2}_{}$$
Center $${}_{T}OK{3}_{}$$

Naar vind de $$x$$ -intercept(s), substitute $$0$$ voor $$y$$ in de cirkel's standaard form vergelijking
$${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$$
en Los op de kwadratische vergelijking voor $$x$$:

$${\left(x-8\right)}^2+{\left(y-12\right)}^2=96$$

$${\left(x-8\right)}^2+{\left(0-12\right)}^2=96$$

$${\left(x-8\right)}^2+{\left(-12\right)}^2=96$$

$${\left(x-8\right)}^2+144=96$$

$${\left(x-8\right)}^2=96-144$$

$${\left(x-8\right)}^2=-48$$

$$\sqrt{\left({\left(x-8\right)}^2\right)}=\sqrt{\left(-48\right)}$$

$$x-8=\sqrt{\left(-48\right)}$$

$$x=\pm \sqrt{\left(-48\right)}+8$$

Nee x-intercepts

Naar vind de $$y$$ -intercept(s), substitute $$0$$ voor $$x$$ in de cirkel's standaard form vergelijking
$${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$$
en Los op de kwadratische vergelijking voor $$y$$:

$${\left(x-8\right)}^2+{\left(y-12\right)}^2=96$$

$${\left(0-8\right)}^2+{\left(y-12\right)}^2=96$$

$${\left(-8\right)}^2+{\left(y-12\right)}^2=96$$

$$64+{\left(y-12\right)}^2=96$$

$${\left(y-12\right)}^2=96-64$$

$${\left(y-12\right)}^2=32$$

$$\sqrt{\left({\left(y-12\right)}^2\right)}=\sqrt{\left(32\right)}$$

$$y-12=\sqrt{\left(32\right)}$$

$$y=\pm \sqrt{\left(32\right)}+12$$

$$y_1=\left(0,\sqrt{\left(32\right)}+12\right),y_2=\left(0,-\sqrt{\left(32\right)}+12\right)$$