已知 ${O_1}$ 和 ${O_2}$ 是平面上两个不重合的固定圆周，$C$ 是平面上的一个动圆且与 ${O_1}$、${O_2}$ 都相切．问：$C$ 的圆心轨迹是何种曲线？证明你的结论．
【难度】
【出处】
2011年北京大学等三校联考自主招生保送生测试
【标注】
【答案】
略
【解析】
设圆 ${O_1}$、圆 ${O_2}$、圆 $C$ 的半径分别为 ${r_1}$、${r_2}$、$r$，且 ${r_1} \leqslant {r_2}$，$\left| {C{O_1}} \right| = {d_1}$，$\left| {C{O_2}} \right| = {d_2}$，则
圆 $C$ 与圆 ${O_1}$ 相切就有 ${d_1} = {r_1} + r$ 或 ${d_1} = {r_1} - r$ 或 ${d_1} = r - {r_1}$，类似的，列出下表：\[\begin{array}{|c|c|c|c|c|}\hline &{d_1} = {r_1} + r&{d_1} = {r_1} - r&{d_1} = r - {r_1}\\ \hline {d_2} = {r_2} + r&{d_2} - {d_1} = {r_2} - {r_1}&{d_1} + {d_2} = {r_1} + {r_2}&{d_2} - {d_1} = {r_1} + {r_2}\\ \hline {d_2} = {r_2} - r&{d_1} + {d_2} = {r_1} + {r_2}&{d_2} - {d_1} = {r_2} - {r_1}&{d_1} + {d_2} = {r_2} - {r_1}\\ \hline {d_2} = r - {r_2}&{d_1} - {d_2} = {r_1} + {r_2}&{d_1} + {d_2} = {r_1} - {r_2}&{d_1} - {d_2} = {r_2} - {r_1}\\ \hline\end{array}\]于是轨迹可能包含这些部分：\[\begin{split}&{G_1}:\left| {{d_1} - {d_2}} \right| = {r_2} - {r_1};\\&{G_2}:\left| {{d_1} - {d_2}} \right| = {r_1} + {r_2};\\&{G_3}:\left| {{d_1} + {d_2}} \right| = {r_2} - {r_1};\\&{G_4}:\left| {{d_1} + {d_2}} \right| = {r_1} + {r_2}.\end{split}\]接下来考虑 $\left| {{O_1}{O_2}} \right| $ $ = d $；
当 $ r_{2}>r_{1}$ 时：$$\begin{array}{|c|c|c|c|c|c|c|}\hline d&0&\left( {0 , {r_2} - {r_1}} \right)&{r_2} - {r_1}&\left( {{r_2} - {r_1} , {r_1} + {r_2}} \right)&{r_1} + {r_2}&\left( {{r_1} + {r_2} , + \infty } \right)\\ \hline {G_1}&\text{不存在}&\text{不存在}&\text{两条射线}&\text{双曲线}&\text{双曲线}&\text{双曲线}\\ \hline {G_2}&\text{不存在}&\text{不存在}&\text{不存在}&\text{不存在}&\text{两条射线}&\text{双曲线}\\ \hline {G_3}&\text{圆}&\text{椭圆}&\text{线段}{O_1}{O_2}&\text{不存在}&\text{不存在}&\text{不存在}\\ \hline {G_4}&\text{圆}&\text{椭圆}&\text{椭圆}&\text{椭圆}&\text{线段}&\text{不存在}\\ \hline \end{array}$$当 $r_{2}=r_{1}$ 时：$$\begin{array}{|c|c|c|c|c|}\hline d&0&\left( {0 , {r_2} +{r_1}} \right)&{r_2} + {r_1}&\left( {{r_1} + {r_2} , + \infty } \right)\\ \hline {G_1}&\text{不符合题意}&{O_1}{O_2}\text{的垂直平分线}&{O_1}{O_2}\text{的垂直平分线}&{O_1}{O_2}\text{的垂直平分线}\\ \hline {G_2}&\text{不符合题意}&\text{不存在}&\text{两条射线}&\text{双曲线}\\ \hline {G_3}&\text{不符合题意}&\text{不存在}&\text{不存在}&\text{不存在}\\ \hline {G_4}&\text{不符合题意}&\text{椭圆}&\text{线段}&\text{不存在}\\ \hline \end{array}$$因此，一共有 ${r_2} > {r_1}$ 时有 $6$ 种情况；${r_2} = {r_1}$ 时有 $3$ 种情况．
圆 $C$ 与圆 ${O_1}$ 相切就有 ${d_1} = {r_1} + r$ 或 ${d_1} = {r_1} - r$ 或 ${d_1} = r - {r_1}$，类似的，列出下表：\[\begin{array}{|c|c|c|c|c|}\hline &{d_1} = {r_1} + r&{d_1} = {r_1} - r&{d_1} = r - {r_1}\\ \hline {d_2} = {r_2} + r&{d_2} - {d_1} = {r_2} - {r_1}&{d_1} + {d_2} = {r_1} + {r_2}&{d_2} - {d_1} = {r_1} + {r_2}\\ \hline {d_2} = {r_2} - r&{d_1} + {d_2} = {r_1} + {r_2}&{d_2} - {d_1} = {r_2} - {r_1}&{d_1} + {d_2} = {r_2} - {r_1}\\ \hline {d_2} = r - {r_2}&{d_1} - {d_2} = {r_1} + {r_2}&{d_1} + {d_2} = {r_1} - {r_2}&{d_1} - {d_2} = {r_2} - {r_1}\\ \hline\end{array}\]于是轨迹可能包含这些部分：\[\begin{split}&{G_1}:\left| {{d_1} - {d_2}} \right| = {r_2} - {r_1};\\&{G_2}:\left| {{d_1} - {d_2}} \right| = {r_1} + {r_2};\\&{G_3}:\left| {{d_1} + {d_2}} \right| = {r_2} - {r_1};\\&{G_4}:\left| {{d_1} + {d_2}} \right| = {r_1} + {r_2}.\end{split}\]接下来考虑 $\left| {{O_1}{O_2}} \right| $ $ = d $；
当 $ r_{2}>r_{1}$ 时：$$\begin{array}{|c|c|c|c|c|c|c|}\hline d&0&\left( {0 , {r_2} - {r_1}} \right)&{r_2} - {r_1}&\left( {{r_2} - {r_1} , {r_1} + {r_2}} \right)&{r_1} + {r_2}&\left( {{r_1} + {r_2} , + \infty } \right)\\ \hline {G_1}&\text{不存在}&\text{不存在}&\text{两条射线}&\text{双曲线}&\text{双曲线}&\text{双曲线}\\ \hline {G_2}&\text{不存在}&\text{不存在}&\text{不存在}&\text{不存在}&\text{两条射线}&\text{双曲线}\\ \hline {G_3}&\text{圆}&\text{椭圆}&\text{线段}{O_1}{O_2}&\text{不存在}&\text{不存在}&\text{不存在}\\ \hline {G_4}&\text{圆}&\text{椭圆}&\text{椭圆}&\text{椭圆}&\text{线段}&\text{不存在}\\ \hline \end{array}$$当 $r_{2}=r_{1}$ 时：$$\begin{array}{|c|c|c|c|c|}\hline d&0&\left( {0 , {r_2} +{r_1}} \right)&{r_2} + {r_1}&\left( {{r_1} + {r_2} , + \infty } \right)\\ \hline {G_1}&\text{不符合题意}&{O_1}{O_2}\text{的垂直平分线}&{O_1}{O_2}\text{的垂直平分线}&{O_1}{O_2}\text{的垂直平分线}\\ \hline {G_2}&\text{不符合题意}&\text{不存在}&\text{两条射线}&\text{双曲线}\\ \hline {G_3}&\text{不符合题意}&\text{不存在}&\text{不存在}&\text{不存在}\\ \hline {G_4}&\text{不符合题意}&\text{椭圆}&\text{线段}&\text{不存在}\\ \hline \end{array}$$因此，一共有 ${r_2} > {r_1}$ 时有 $6$ 种情况；${r_2} = {r_1}$ 时有 $3$ 种情况．
答案
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