解答 - 圆的性质

直径$$\left(d\right)$$等于半径的两倍：
$$d=2·r$$

$$d=2\cdot r$$

r=2.449

$$d=2\cdot 2.449$$

$$d=4.899$$

周长$$\left(c\right)$$等于半径的两倍乘以π：
$$c=2·r·\pi$$

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

r=2.449

$$c=2\cdot 2.449\cdot \pi$$

$$c=4.899\cdot \pi$$

面积$$\left(a\right)$$等于半径的平方乘以π：
$$a=r^2·\pi$$

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

r=2.449

$$a={2.449}^2\cdot \pi$$

$$a=6\cdot \pi$$

圆心的坐标通常由$$h$$和$$k$$在圆的标准形式方程中表示：
$${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$$
在方程中确定$$h$$和$$k$$：
$$i^2+{\left(y+3\right)}^2=6$$
$$h=0$$
$$k=-3$$
Center $$\left(0;-3\right)$$

为了找到$$x$$的交点，将$$0$$代入$$y$$到圆的标准形式方程
$${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$$
并为$$x$$解二次方程式：

$${\left(i+0\right)}^2+{\left(y+3\right)}^2=6$$

$${\left(i+0\right)}^2+{\left(0+3\right)}^2=6$$

$${\left(i+0\right)}^2+{\left(3\right)}^2=6$$

$${\left(i+0\right)}^2+9=6$$

$${\left(i+0\right)}^2=6-9$$

$${\left(i+0\right)}^2=-3$$

$$\sqrt{\left({\left(i+0\right)}^2\right)}=\sqrt{\left(-3\right)}$$

$$i+0=\sqrt{\left(-3\right)}$$

$$i=\pm \sqrt{\left(-3\right)}-0$$

无x截距

求解$$y$$截距，将$$0$$代入圆方程标准形式$${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$$
然后解该二次方程获得$$y$$：

$${\left(i+0\right)}^2+{\left(y+3\right)}^2=6$$

$${\left(0+0\right)}^2+{\left(y+3\right)}^2=6$$

$${\left(-0\right)}^2+{\left(y+3\right)}^2=6$$

$$0+{\left(y+3\right)}^2=6$$

$${\left(y+3\right)}^2=6-0$$

$${\left(y+3\right)}^2=6$$

$$\sqrt{\left({\left(y+3\right)}^2\right)}=\sqrt{\left(6\right)}$$

$$y+3=\sqrt{\left(6\right)}$$

$$y=\pm \sqrt{\left(6\right)}-3$$

$$y_1=\left(0,\sqrt{\left(6\right)}-3\right),y_2=\left(0,-\sqrt{\left(6\right)}-3\right)$$

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