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=4

$$d=2\cdot 4$$

$$d=8$$

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=4

$$c=2\cdot 4\cdot \pi$$

$$c=8\cdot \pi$$

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=4

$$a=4^2\cdot \pi$$

$$a=16\cdot \pi$$

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$$k$$：
$${\left(x-1\right)}^2+{\left(y+8\right)}^2=16$$
$$h=1$$
$$k=-8$$
Center $$\left(1;-8\right)$$

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：

$${\left(x-1\right)}^2+{\left(y+8\right)}^2=16$$

$${\left(x-1\right)}^2+{\left(0+8\right)}^2=16$$

$${\left(x-1\right)}^2+{\left(8\right)}^2=16$$

$${\left(x-1\right)}^2+64=16$$

$${\left(x-1\right)}^2=16-64$$

$${\left(x-1\right)}^2=-48$$

$$\sqrt{\left({\left(x-1\right)}^2\right)}=\sqrt{\left(-48\right)}$$

$$x-1=\sqrt{\left(-48\right)}$$

$$x=\pm \sqrt{\left(-48\right)}+1$$

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$$y$$：

$${\left(x-1\right)}^2+{\left(y+8\right)}^2=16$$

$${\left(0-1\right)}^2+{\left(y+8\right)}^2=16$$

$${\left(-1\right)}^2+{\left(y+8\right)}^2=16$$

$$1+{\left(y+8\right)}^2=16$$

$${\left(y+8\right)}^2=16-1$$

$${\left(y+8\right)}^2=15$$

$$\sqrt{\left({\left(y+8\right)}^2\right)}=\sqrt{\left(15\right)}$$

$$y+8=\sqrt{\left(15\right)}$$

$$y=\pm \sqrt{\left(15\right)}-8$$

$$y_1=\left(0,\sqrt{\left(15\right)}-8\right),y_2=\left(0,-\sqrt{\left(15\right)}-8\right)$$

CircleFromEquationSolverStep7TextUnit1