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=60.117

$$d=2\cdot 60.117$$

$$d=120.233$$

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=60.117

$$c=2\cdot 60.117\cdot \pi$$

$$c=120.233\cdot \pi$$

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=60.117

$$a={60.117}^2\cdot \pi$$

$$a=3614\cdot \pi$$

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$$k$$：
$$x+{233}^2+y-{253}^2=3614$$
$$h=-233$$
$$k=253$$
Center $$\left(-233;253\right)$$

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：

$${\left(x+233\right)}^2+{\left(y-253\right)}^2=3614$$

$${\left(x+233\right)}^2+{\left(0-253\right)}^2=3614$$

$${\left(x+233\right)}^2+{\left(-253\right)}^2=3614$$

$${\left(x+233\right)}^2+64009=3614$$

$${\left(x+233\right)}^2=3614-64009$$

$${\left(x+233\right)}^2=-60395$$

$$\sqrt{\left({\left(x+233\right)}^2\right)}=\sqrt{\left(-60395\right)}$$

$$x+233=\sqrt{\left(-60395\right)}$$

$$x=\pm \sqrt{\left(-60395\right)}-233$$

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$$y$$：

$${\left(x+233\right)}^2+{\left(y-253\right)}^2=3614$$

$${\left(0+233\right)}^2+{\left(y-253\right)}^2=3614$$

$${\left(233\right)}^2+{\left(y-253\right)}^2=3614$$

$$54289+{\left(y-253\right)}^2=3614$$

$${\left(y-253\right)}^2=3614-54289$$

$${\left(y-253\right)}^2=-50675$$

$$\sqrt{\left({\left(y-253\right)}^2\right)}=\sqrt{\left(-50675\right)}$$

$$y-253=\sqrt{\left(-50675\right)}$$

$$y=\pm \sqrt{\left(-50675\right)}+253$$

无y截距

CircleFromEquationSolverStep7TextUnit1