Rešenje - Osobine kružnica

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

$$d=2\cdot r$$

r=6

$$d=2\cdot 6$$

$$d=12$$

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

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

r=6

$$c=2\cdot 6\cdot \pi$$

$$c=12\cdot \pi$$

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

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

r=6

$$a=6^2\cdot \pi$$

$$a=36\cdot \pi$$

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:
$${\left(x+2\right)}^2+{\left(y-1\right)}^2=36$$
$$h=-2$$
$$k=1$$
Središte $$\left(-2;1\right)$$

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 $$x$$:

$${\left(x+2\right)}^2+{\left(y-1\right)}^2=36$$

$${\left(x+2\right)}^2+{\left(0-1\right)}^2=36$$

$${\left(x+2\right)}^2+{\left(-1\right)}^2=36$$

$${\left(x+2\right)}^2+1=36$$

$${\left(x+2\right)}^2=36-1$$

$${\left(x+2\right)}^2=35$$

$$\sqrt{\left({\left(x+2\right)}^2\right)}=\sqrt{\left(35\right)}$$

$$x+2=\sqrt{\left(35\right)}$$

$$x=\pm \sqrt{\left(35\right)}-2$$

$$x_1=\left(\sqrt{\left(35\right)}-2,0\right),x_2=\left(-\sqrt{\left(35\right)}-2,0\right)$$

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 $$y$$:

$${\left(x+2\right)}^2+{\left(y-1\right)}^2=36$$

$${\left(0+2\right)}^2+{\left(y-1\right)}^2=36$$

$${\left(2\right)}^2+{\left(y-1\right)}^2=36$$

$$4+{\left(y-1\right)}^2=36$$

$${\left(y-1\right)}^2=36-4$$

$${\left(y-1\right)}^2=32$$

$$\sqrt{\left({\left(y-1\right)}^2\right)}=\sqrt{\left(32\right)}$$

$$y-1=\sqrt{\left(32\right)}$$

$$y=\pm \sqrt{\left(32\right)}+1$$

$$y_1=\left(0,\sqrt{\left(32\right)}+1\right),y_2=\left(0,-\sqrt{\left(32\right)}+1\right)$$