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=19.105

$$d=2\cdot 19.105$$

$$d=38.21$$

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=19.105

$$c=2\cdot 19.105\cdot \pi$$

$$c=38.21\cdot \pi$$

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=19.105

$$a={19.105}^2\cdot \pi$$

$$a=365\cdot \pi$$

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$$k$$：
$$x^2+{\left(n+1\right)}^2=365$$
$$h=0$$
$$k=-1$$
Center $$\left(0;-1\right)$$

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：

$${\left(x+0\right)}^2+{\left(n+1\right)}^2=365$$

$${\left(x+0\right)}^2+{\left(0+1\right)}^2=365$$

$${\left(x+0\right)}^2+{\left(1\right)}^2=365$$

$${\left(x+0\right)}^2+1=365$$

$${\left(x+0\right)}^2=365-1$$

$${\left(x+0\right)}^2=364$$

$$\sqrt{\left({\left(x+0\right)}^2\right)}=\sqrt{\left(364\right)}$$

$$x+0=\sqrt{\left(364\right)}$$

$$x=\pm \sqrt{\left(364\right)}-0$$

$$x_1=\left(\sqrt{\left(364\right)}-0,0\right),x_2=\left(-\sqrt{\left(364\right)}-0,0\right)$$

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$$y$$：

$${\left(x+0\right)}^2+{\left(n+1\right)}^2=365$$

$${\left(0+0\right)}^2+{\left(n+1\right)}^2=365$$

$${\left(-0\right)}^2+{\left(n+1\right)}^2=365$$

$$0+{\left(n+1\right)}^2=365$$

$${\left(n+1\right)}^2=365-0$$

$${\left(n+1\right)}^2=365$$

$$\sqrt{\left({\left(n+1\right)}^2\right)}=\sqrt{\left(365\right)}$$

$$n+1=\sqrt{\left(365\right)}$$

$$n=\pm \sqrt{\left(365\right)}-1$$

$$n_1=\left(0,\sqrt{\left(365\right)}-1\right),n_2=\left(0,-\sqrt{\left(365\right)}-1\right)$$

CircleFromEquationSolverStep7TextUnit1