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=4456.679

$$d=2\cdot 4456.679$$

$$d=8913.358$$

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=4456.679

$$c=2\cdot 4456.679\cdot \pi$$

$$c=8913.358\cdot \pi$$

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

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

r=4456.679

$$a={4456.679}^2\cdot \pi$$

$$a=19861986\cdot \pi$$

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:
$$x^2+y^2=19861986$$
$$h=0$$
$$k=0$$
Sentro $$\left(0;0\right)$$

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 $$x$$:

$${\left(x+0\right)}^2+{\left(y+0\right)}^2=19861986$$

$${\left(x+0\right)}^2+{\left(0+0\right)}^2=19861986$$

$${\left(x+0\right)}^2+{\left(-0\right)}^2=19861986$$

$${\left(x+0\right)}^2+0=19861986$$

$${\left(x+0\right)}^2=19861986-0$$

$${\left(x+0\right)}^2=19861986$$

$$\sqrt{\left({\left(x+0\right)}^2\right)}=\sqrt{\left(19861986\right)}$$

$$x+0=\sqrt{\left(19861986\right)}$$

$$x=\pm \sqrt{\left(19861986\right)}-0$$

$$x_1=\left(\sqrt{\left(19861986\right)}-0,0\right),x_2=\left(-\sqrt{\left(19861986\right)}-0,0\right)$$

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 $$y$$:

$${\left(x+0\right)}^2+{\left(y+0\right)}^2=19861986$$

$${\left(0+0\right)}^2+{\left(y+0\right)}^2=19861986$$

$${\left(-0\right)}^2+{\left(y+0\right)}^2=19861986$$

$$0+{\left(y+0\right)}^2=19861986$$

$${\left(y+0\right)}^2=19861986-0$$

$${\left(y+0\right)}^2=19861986$$

$$\sqrt{\left({\left(y+0\right)}^2\right)}=\sqrt{\left(19861986\right)}$$

$$y+0=\sqrt{\left(19861986\right)}$$

$$y=\pm \sqrt{\left(19861986\right)}-0$$

$$y_1=\left(0,\sqrt{\left(19861986\right)}-0\right),y_2=\left(0,-\sqrt{\left(19861986\right)}-0\right)$$

CircleFromEquationSolverStep7TextUnit1