Solusyon - Katangian ng mga bilog

Ang diameter $$\left(d\right)$$ ay katumbas ng dalawang beses ang radius:
$$d=2·r$$

$$d=2\cdot r$$

r=26

$$d=2\cdot 26$$

$$d=52$$

Ang circumference $$\left(c\right)$$ ay katumbas ng dalawang beses ang radius times π:
$$c=2·r·\pi$$

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

r=26

$$c=2\cdot 26\cdot \pi$$

$$c=52\cdot \pi$$

Ang area $$\left(a\right)$$ ay katumbas ng squared na radius times π:
$$a=r^2·\pi$$

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

r=26

$$a={26}^2\cdot \pi$$

$$a=676\cdot \pi$$

Ang mga coordinates ng sentro ng isang bilog ay karaniwan, ngunit hindi palaging, kinakatawan ni $$h$$ at $$k$$ sa standard form equation ng isang bilog:
$${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$$
Kilalanin ang $$h$$ at $$k$$ sa equation:
$${\left(x-4\right)}^2+{\left(y-2\right)}^2=676$$
$$h=4$$
$$k=2$$
Sentro $$\left(4;2\right)$$

Upang mahanap ang $$x$$ -intercept(s), palitan ng $$0$$ ang $$y$$ sa standard form equation ng bilog
$${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$$
at lutasin ang quadratic equation para sa $$x$$:

$${\left(x-4\right)}^2+{\left(y-2\right)}^2=676$$

$${\left(x-4\right)}^2+{\left(0-2\right)}^2=676$$

$${\left(x-4\right)}^2+{\left(-2\right)}^2=676$$

$${\left(x-4\right)}^2+4=676$$

$${\left(x-4\right)}^2=676-4$$

$${\left(x-4\right)}^2=672$$

$$\sqrt{\left({\left(x-4\right)}^2\right)}=\sqrt{\left(672\right)}$$

$$x-4=\sqrt{\left(672\right)}$$

$$x=\pm \sqrt{\left(672\right)}+4$$

$$x_1=\left(\sqrt{\left(672\right)}+4,0\right),x_2=\left(-\sqrt{\left(672\right)}+4,0\right)$$

Para hanapin ang $$y$$ -intercept(s), palitan ang $$0$$ para sa $$x$$ sa standard form equation ng bilog
$${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$$
at isulit ang quadratic equation para sa $$y$$:

$${\left(x-4\right)}^2+{\left(y-2\right)}^2=676$$

$${\left(0-4\right)}^2+{\left(y-2\right)}^2=676$$

$${\left(-4\right)}^2+{\left(y-2\right)}^2=676$$

$$16+{\left(y-2\right)}^2=676$$

$${\left(y-2\right)}^2=676-16$$

$${\left(y-2\right)}^2=660$$

$$\sqrt{\left({\left(y-2\right)}^2\right)}=\sqrt{\left(660\right)}$$

$$y-2=\sqrt{\left(660\right)}$$

$$y=\pm \sqrt{\left(660\right)}+2$$

$$y_1=\left(0,\sqrt{\left(660\right)}+2\right),y_2=\left(0,-\sqrt{\left(660\right)}+2\right)$$

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