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=7

$$d=2\cdot 7$$

$$d=14$$

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=7

$$c=2\cdot 7\cdot \pi$$

$$c=14\cdot \pi$$

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=7

$$a=7^2\cdot \pi$$

$$a=49\cdot \pi$$

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$$k$$：
$${\left(x-10\right)}^2+{\left(y+6\right)}^2=49$$
$$h=10$$
$$k=-6$$
Center $$\left(10;-6\right)$$

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：

$${\left(x-10\right)}^2+{\left(y+6\right)}^2=49$$

$${\left(x-10\right)}^2+{\left(0+6\right)}^2=49$$

$${\left(x-10\right)}^2+{\left(6\right)}^2=49$$

$${\left(x-10\right)}^2+36=49$$

$${\left(x-10\right)}^2=49-36$$

$${\left(x-10\right)}^2=13$$

$$\sqrt{\left({\left(x-10\right)}^2\right)}=\sqrt{\left(13\right)}$$

$$x-10=\sqrt{\left(13\right)}$$

$$x=\pm \sqrt{\left(13\right)}+10$$

$$x_1=\left(\sqrt{\left(13\right)}+10,0\right),x_2=\left(-\sqrt{\left(13\right)}+10,0\right)$$

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$$y$$：

$${\left(x-10\right)}^2+{\left(y+6\right)}^2=49$$

$${\left(0-10\right)}^2+{\left(y+6\right)}^2=49$$

$${\left(-10\right)}^2+{\left(y+6\right)}^2=49$$

$$100+{\left(y+6\right)}^2=49$$

$${\left(y+6\right)}^2=49-100$$

$${\left(y+6\right)}^2=-51$$

$$\sqrt{\left({\left(y+6\right)}^2\right)}=\sqrt{\left(-51\right)}$$

$$y+6=\sqrt{\left(-51\right)}$$

$$y=\pm \sqrt{\left(-51\right)}-6$$

无y截距

CircleFromEquationSolverStep7TextUnit1