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=3.742

$$d=2\cdot 3.742$$

$$d=7.483$$

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=3.742

$$c=2\cdot 3.742\cdot \pi$$

$$c=7.483\cdot \pi$$

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=3.742

$$a={3.742}^2\cdot \pi$$

$$a=14\cdot \pi$$

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$$k$$：
$${\left(x-6\right)}^2+{\left(y+0\right)}^2=14$$
$$h=6$$
$$k=0$$
Center $$\left(6;0\right)$$

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：

$${\left(x-6\right)}^2+{\left(y+0\right)}^2=14$$

$${\left(x-6\right)}^2+{\left(0+0\right)}^2=14$$

$${\left(x-6\right)}^2+{\left(-0\right)}^2=14$$

$${\left(x-6\right)}^2+0=14$$

$${\left(x-6\right)}^2=14-0$$

$${\left(x-6\right)}^2=14$$

$$\sqrt{\left({\left(x-6\right)}^2\right)}=\sqrt{\left(14\right)}$$

$$x-6=\sqrt{\left(14\right)}$$

$$x=\pm \sqrt{\left(14\right)}+6$$

$$x_1=\left(\sqrt{\left(14\right)}+6,0\right),x_2=\left(-\sqrt{\left(14\right)}+6,0\right)$$

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$$y$$：

$${\left(x-6\right)}^2+{\left(y+0\right)}^2=14$$

$${\left(0-6\right)}^2+{\left(y+0\right)}^2=14$$

$${\left(-6\right)}^2+{\left(y+0\right)}^2=14$$

$$36+{\left(y+0\right)}^2=14$$

$${\left(y+0\right)}^2=14-36$$

$${\left(y+0\right)}^2=-22$$

$$\sqrt{\left({\left(y+0\right)}^2\right)}=\sqrt{\left(-22\right)}$$

$$y+0=\sqrt{\left(-22\right)}$$

$$y=\pm \sqrt{\left(-22\right)}-0$$

无y截距

CircleFromEquationSolverStep7TextUnit1