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=0

$$d=2\cdot 0$$

$$d=0$$

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=0

$$c=2\cdot 0\cdot \pi$$

$$c=0\cdot \pi$$

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=0

$$a=0^2\cdot \pi$$

$$a=0\cdot \pi$$

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

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：

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

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

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

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

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

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

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

$$y+2=\sqrt{\left(-1\right)}$$

$$y=\pm \sqrt{\left(-1\right)}-2$$

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$$y$$：

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

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

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

$$4+{\left(x-1\right)}^2=0$$

$${\left(x-1\right)}^2=0-4$$

$${\left(x-1\right)}^2=-4$$

$$\sqrt{\left({\left(x-1\right)}^2\right)}=\sqrt{\left(-4\right)}$$

$$x-1=\sqrt{\left(-4\right)}$$

$$x=\pm \sqrt{\left(-4\right)}+1$$

无y截距

CircleFromEquationSolverStep7TextUnit1