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=4.243

$$d=2\cdot 4.243$$

$$d=8.485$$

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=4.243

$$c=2\cdot 4.243\cdot \pi$$

$$c=8.485\cdot \pi$$

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=4.243

$$a={4.243}^2\cdot \pi$$

$$a=18\cdot \pi$$

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$$k$$：
$$x-{14}^2+y-{3346}^2=18$$
$$h=14$$
$$k=3346$$
Center $$\left(14;3346\right)$$

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：

$${\left(x-14\right)}^2+{\left(y-3346\right)}^2=18$$

$${\left(x-14\right)}^2+{\left(0-3346\right)}^2=18$$

$${\left(x-14\right)}^2+{\left(-3346\right)}^2=18$$

$${\left(x-14\right)}^2+11195716=18$$

$${\left(x-14\right)}^2=18-11195716$$

$${\left(x-14\right)}^2=-11195698$$

$$\sqrt{\left({\left(x-14\right)}^2\right)}=\sqrt{\left(-11195698\right)}$$

$$x-14=\sqrt{\left(-11195698\right)}$$

$$x=\pm \sqrt{\left(-11195698\right)}+14$$

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$$y$$：

$${\left(x-14\right)}^2+{\left(y-3346\right)}^2=18$$

$${\left(0-14\right)}^2+{\left(y-3346\right)}^2=18$$

$${\left(-14\right)}^2+{\left(y-3346\right)}^2=18$$

$$196+{\left(y-3346\right)}^2=18$$

$${\left(y-3346\right)}^2=18-196$$

$${\left(y-3346\right)}^2=-178$$

$$\sqrt{\left({\left(y-3346\right)}^2\right)}=\sqrt{\left(-178\right)}$$

$$y-3346=\sqrt{\left(-178\right)}$$

$$y=\pm \sqrt{\left(-178\right)}+3346$$

无y截距

CircleFromEquationSolverStep7TextUnit1