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=7.874

$$d=2\cdot 7.874$$

$$d=15.748$$

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=7.874

$$c=2\cdot 7.874\cdot \pi$$

$$c=15.748\cdot \pi$$

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=7.874

$$a={7.874}^2\cdot \pi$$

$$a=62\cdot \pi$$

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$$k$$：
$$t^2+d^2=62$$
$$h=0$$
$$k=0$$
Center $$\left(0;0\right)$$

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：

$${\left(t+0\right)}^2+{\left(d+0\right)}^2=62$$

$${\left(t+0\right)}^2+{\left(0+0\right)}^2=62$$

$${\left(t+0\right)}^2+{\left(-0\right)}^2=62$$

$${\left(t+0\right)}^2+0=62$$

$${\left(t+0\right)}^2=62-0$$

$${\left(t+0\right)}^2=62$$

$$\sqrt{\left({\left(t+0\right)}^2\right)}=\sqrt{\left(62\right)}$$

$$t+0=\sqrt{\left(62\right)}$$

$$t=\pm \sqrt{\left(62\right)}-0$$

$$t_1=\left(\sqrt{\left(62\right)}-0,0\right),t_2=\left(-\sqrt{\left(62\right)}-0,0\right)$$

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$$y$$：

$${\left(t+0\right)}^2+{\left(d+0\right)}^2=62$$

$${\left(0+0\right)}^2+{\left(d+0\right)}^2=62$$

$${\left(-0\right)}^2+{\left(d+0\right)}^2=62$$

$$0+{\left(d+0\right)}^2=62$$

$${\left(d+0\right)}^2=62-0$$

$${\left(d+0\right)}^2=62$$

$$\sqrt{\left({\left(d+0\right)}^2\right)}=\sqrt{\left(62\right)}$$

$$d+0=\sqrt{\left(62\right)}$$

$$d=\pm \sqrt{\left(62\right)}-0$$

$$d_1=\left(0,\sqrt{\left(62\right)}-0\right),d_2=\left(0,-\sqrt{\left(62\right)}-0\right)$$

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