Rešenje - Osobine kružnica

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

$$d=2\cdot r$$

r=10,295630140987

$$d=2\cdot 10,295630140987$$

$$d=20,591260281974$$

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

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

r=10,295630140987

$$c=2\cdot 10,295630140987\cdot \pi$$

$$c=20,591260281974\cdot \pi$$

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

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

r=10,295630140987

$$a=10,{295630140987}^2\cdot \pi$$

$$a=106\cdot \pi$$

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:
$${\left(x-8\right)}^2+{\left(y+3\right)}^2=106$$
$$h=8$$
$$k=-3$$
Središte $$\left(8;-3\right)$$

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 $$x$$:

$${\left(x-8\right)}^2+{\left(y+3\right)}^2=106$$

$${\left(x-8\right)}^2+{\left(0+3\right)}^2=106$$

$${\left(x-8\right)}^2+{\left(3\right)}^2=106$$

$${\left(x-8\right)}^2+9=106$$

$${\left(x-8\right)}^2=106-9$$

$${\left(x-8\right)}^2=97$$

$$\sqrt{\left({\left(x-8\right)}^2\right)}=\sqrt{\left(97\right)}$$

$$x-8=\sqrt{\left(97\right)}$$

$$x=\pm \sqrt{\left(97\right)}+8$$

$$x_1=\left(\sqrt{\left(97\right)}+8,0\right),x_2=\left(-\sqrt{\left(97\right)}+8,0\right)$$

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 $$y$$:

$${\left(x-8\right)}^2+{\left(y+3\right)}^2=106$$

$${\left(0-8\right)}^2+{\left(y+3\right)}^2=106$$

$${\left(-8\right)}^2+{\left(y+3\right)}^2=106$$

$$64+{\left(y+3\right)}^2=106$$

$${\left(y+3\right)}^2=106-64$$

$${\left(y+3\right)}^2=42$$

$$\sqrt{\left({\left(y+3\right)}^2\right)}=\sqrt{\left(42\right)}$$

$$y+3=\sqrt{\left(42\right)}$$

$$y=\pm \sqrt{\left(42\right)}-3$$

$$y_1=\left(0,\sqrt{\left(42\right)}-3\right),y_2=\left(0,-\sqrt{\left(42\right)}-3\right)$$