Rešenje - Osobine kružnica

Prečnik $$\left(d\right)$$ ima dvostruku vrednost radijusa:

$$d=2\cdot r$$

r=77,61443164772902

$$d=2\cdot 77,61443164772902$$

$$d=155,22886329545804$$

Opseg $$\left(c\right)$$ ima dvostruku vrednost radijusa pomnoženo sa π:

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

r=77,61443164772902

$$c=2\cdot 77,61443164772902\cdot \pi$$

$$c=155,22886329545804\cdot \pi$$

Površina $$\left(a\right)$$ ima vrednost kvadrata radijusa pomnoženo sa π:

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

r=77,61443164772902

$$a=77,{61443164772902}^2\cdot \pi$$

$$a=6024\cdot \pi$$

Koordinate središta kružnice se obično, ali ne uvek, prikazuju sa $$h$$ i $$k$$ u standardnoj jednačini kružnice: $${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$$
Otkrijte $$h$$ i $$k$$ u jednačini:
$$y^2+z^2=6024$$
$$h=0$$
$$k=0$$
Središte $$\left(0;0\right)$$

Da biste našli $$x$$ -presek(e), supstitucirajte $$0$$ za $$y$$ u standardnoj formi jednačine kruga
$${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$$
i rešite kvadratnu jednačinu za $$x$$:

$${\left(y+0\right)}^2+{\left(z+0\right)}^2=6024$$

$${\left(y+0\right)}^2+{\left(0+0\right)}^2=6024$$

$${\left(y+0\right)}^2+{\left(-0\right)}^2=6024$$

$${\left(y+0\right)}^2+0=6024$$

$${\left(y+0\right)}^2=6024-0$$

$${\left(y+0\right)}^2=6024$$

$$\sqrt{\left({\left(y+0\right)}^2\right)}=\sqrt{\left(6024\right)}$$

$$y+0=\sqrt{\left(6024\right)}$$

$$y=\pm \sqrt{\left(6024\right)}-0$$

$$y_1=\left(\sqrt{\left(6024\right)}-0,0\right),y_2=\left(-\sqrt{\left(6024\right)}-0,0\right)$$

Da bi se pronašli odsečci na osi $$y$$, zameni $$0$$ za $$x$$ u standardnoj jednačini kružnice $${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$$ i reši kvadratnu jednačinu za $$y$$:

$${\left(y+0\right)}^2+{\left(z+0\right)}^2=6024$$

$${\left(0+0\right)}^2+{\left(z+0\right)}^2=6024$$

$${\left(-0\right)}^2+{\left(z+0\right)}^2=6024$$

$$0+{\left(z+0\right)}^2=6024$$

$${\left(z+0\right)}^2=6024-0$$

$${\left(z+0\right)}^2=6024$$

$$\sqrt{\left({\left(z+0\right)}^2\right)}=\sqrt{\left(6024\right)}$$

$$z+0=\sqrt{\left(6024\right)}$$

$$z=\pm \sqrt{\left(6024\right)}-0$$

$$z_1=\left(0,\sqrt{\left(6024\right)}-0\right),z_2=\left(0,-\sqrt{\left(6024\right)}-0\right)$$