Rešenje - Osobine kružnica

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

$$d=2\cdot r$$

r=3,872983346207417

$$d=2\cdot 3,872983346207417$$

$$d=7,745966692414834$$

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

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

r=3,872983346207417

$$c=2\cdot 3,872983346207417\cdot \pi$$

$$c=7,745966692414834\cdot \pi$$

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

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

r=3,872983346207417

$$a=3,{872983346207417}^2\cdot \pi$$

$$a=15\cdot \pi$$

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:
$${\left(x-1\right)}^2+{\left(y+3\right)}^2=15$$
$$h=1$$
$$k=-3$$
Središte $$\left(1;-3\right)$$

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 $$x$$:

$${\left(x-1\right)}^2+{\left(y+3\right)}^2=15$$

$${\left(x-1\right)}^2+{\left(0+3\right)}^2=15$$

$${\left(x-1\right)}^2+{\left(3\right)}^2=15$$

$${\left(x-1\right)}^2+9=15$$

$${\left(x-1\right)}^2=15-9$$

$${\left(x-1\right)}^2=6$$

$$\sqrt{\left({\left(x-1\right)}^2\right)}=\sqrt{\left(6\right)}$$

$$x-1=\sqrt{\left(6\right)}$$

$$x=\pm \sqrt{\left(6\right)}+1$$

$$x_1=\left(\sqrt{\left(6\right)}+1,0\right),x_2=\left(-\sqrt{\left(6\right)}+1,0\right)$$

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 $$y$$:

$${\left(x-1\right)}^2+{\left(y+3\right)}^2=15$$

$${\left(0-1\right)}^2+{\left(y+3\right)}^2=15$$

$${\left(-1\right)}^2+{\left(y+3\right)}^2=15$$

$$1+{\left(y+3\right)}^2=15$$

$${\left(y+3\right)}^2=15-1$$

$${\left(y+3\right)}^2=14$$

$$\sqrt{\left({\left(y+3\right)}^2\right)}=\sqrt{\left(14\right)}$$

$$y+3=\sqrt{\left(14\right)}$$

$$y=\pm \sqrt{\left(14\right)}-3$$

$$y_1=\left(0,\sqrt{\left(14\right)}-3\right),y_2=\left(0,-\sqrt{\left(14\right)}-3\right)$$