Solusi - Sifat-sifat lingkaran

Diameter $$\left(d\right)$$ sama dengan dua kali jari-jari:

$$d=2\cdot r$$

r=25

$$d=2\cdot 25$$

$$d=50$$

Keliling $$\left(c\right)$$ sama dengan dua kali jari-jari dikali π:

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

r=25

$$c=2\cdot 25\cdot \pi$$

$$c=50\cdot \pi$$

Luas $$\left(a\right)$$ sama dengan kuadrat jari-jari dikali π:

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

r=25

$$a={25}^2\cdot \pi$$

$$a=625\cdot \pi$$

Koordinat pusat lingkaran biasanya, tetapi tidak selalu, diwakili oleh $$h$$ dan $$k$$ dalam persamaan bentuk umum lingkaran: $${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$$
Tentukan $$h$$ dan $$k$$ dalam persamaan:
$$x^2+y^2=625$$
$$h=0$$
$$k=0$$
Pusat $$\left(0;0\right)$$

To find the $$x$$ -intercept(s), substitute $$0$$ for $$y$$ in the circle's standard form equation
$${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$$
and solve the quadratic equation for $$x$$:

$${\left(x+0\right)}^2+{\left(y+0\right)}^2=625$$

$${\left(x+0\right)}^2+{\left(0+0\right)}^2=625$$

$${\left(x+0\right)}^2+{\left(-0\right)}^2=625$$

$${\left(x+0\right)}^2+0=625$$

$${\left(x+0\right)}^2=625-0$$

$${\left(x+0\right)}^2=625$$

$$\sqrt{\left({\left(x+0\right)}^2\right)}=\sqrt{\left(625\right)}$$

$$x+0=\sqrt{\left(625\right)}$$

$$x=\pm \sqrt{\left(625\right)}-0$$

$$x=\pm 25-0$$

$$x_1=\left(-25;0\right),x_2=\left(25;0\right)$$

Untuk menentukan perpotongan sumbu $$y$$, substitusikan $$x$$ dengan $$0$$ dalam persamaan bentuk standar lingkaran $${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$$ dan pecahkan soal persamaan kuadrat untuk $$y$$:

$${\left(x+0\right)}^2+{\left(y+0\right)}^2=625$$

$${\left(0+0\right)}^2+{\left(y+0\right)}^2=625$$

$${\left(-0\right)}^2+{\left(y+0\right)}^2=625$$

$$0+{\left(y+0\right)}^2=625$$

$${\left(y+0\right)}^2=625-0$$

$${\left(y+0\right)}^2=625$$

$$\sqrt{\left({\left(y+0\right)}^2\right)}=\sqrt{\left(625\right)}$$

$$y+0=\sqrt{\left(625\right)}$$

$$y=\pm \sqrt{\left(625\right)}-0$$

$$y=\pm 25-0$$

$$y_1=\left(0;-25\right),y_2=\left(0;25\right)$$