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=6

$$d=2\cdot 6$$

$$d=12$$

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=6

$$c=2\cdot 6\cdot \pi$$

$$c=12\cdot \pi$$

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=6

$$a=6^2\cdot \pi$$

$$a=36\cdot \pi$$

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$$k$$：
$${\left(x-8\right)}^2+y^2=36$$
$$h=8$$
$$k=0$$
Center $$\left(8;0\right)$$

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：

$${\left(x-8\right)}^2+{\left(y+0\right)}^2=36$$

$${\left(x-8\right)}^2+{\left(0+0\right)}^2=36$$

$${\left(x-8\right)}^2+{\left(-0\right)}^2=36$$

$${\left(x-8\right)}^2+0=36$$

$${\left(x-8\right)}^2=36-0$$

$${\left(x-8\right)}^2=36$$

$$\sqrt{\left({\left(x-8\right)}^2\right)}=\sqrt{\left(36\right)}$$

$$x-8=\sqrt{\left(36\right)}$$

$$x=\pm \sqrt{\left(36\right)}+8$$

$$x=\pm 6+8$$

$$x_1=\left(2;0\right),x_2=\left(14;0\right)$$

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$$y$$：

$${\left(x-8\right)}^2+{\left(y+0\right)}^2=36$$

$${\left(0-8\right)}^2+{\left(y+0\right)}^2=36$$

$${\left(-8\right)}^2+{\left(y+0\right)}^2=36$$

$$64+{\left(y+0\right)}^2=36$$

$${\left(y+0\right)}^2=36-64$$

$${\left(y+0\right)}^2=-28$$

$$\sqrt{\left({\left(y+0\right)}^2\right)}=\sqrt{\left(-28\right)}$$

$$y+0=\sqrt{\left(-28\right)}$$

$$y=\pm \sqrt{\left(-28\right)}-0$$

无y截距

CircleFromEquationSolverStep7TextUnit1