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            • 1.

              在\(\Delta ABC\)中,\(B\left( -2,0 \right),C\left( 2,0 \right),A\left( x,y \right)\),给出\(\Delta ABC\)满足的条件,就能得到动点\(A\)的轨迹方程,下表给出了一些条件及方程:

              条件

              \(①\) \(\Delta ABC\) 周长为 \(10\)

              \(②\) \(\Delta ABC\) 面积为 \(10\)

              \(③\) \(\Delta ABC\) 中, \(\angle A={{90}^{\circ }}\)

              方程

              \({{C}_{1}}:{{y}^{2}}=25\)

              \({{C}_{2}}:{{x}^{2}}+{{y}^{2}}=4\left( y\ne 0 \right)\)

              \({{C}_{3}}:\dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{5}=1\left( y\ne 0 \right)\)

              则满足条件\(①\),\(②\),\(③\)的轨迹方程依次为\((\)   \()\)

              A.\({{C}_{1}},{{C}_{2}},{{C}_{3}}\)
              B.\({{C}_{3}},{{C}_{1}},{{C}_{2}}\)
              C.\({{C}_{3}},{{C}_{2}},{{C}_{1}}\)
              D.\({{C}_{1}},{{C}_{3}},{{C}_{2}}\)
            • 2.

              动圆\(M\)与圆\({{C}_{1}}:{{\left( x+1 \right)}^{2}}+{{y}^{2}}=1\)外切,与圆\({{C}_{2}}:{{\left( x-1 \right)}^{2}}+{{y}^{2}}=25\)内切,则动圆圆心\(M\)的轨迹方程是(    )

              A.\(\dfrac{{{x}^{2}}}{8}+\dfrac{{{y}^{2}}}{9}=1\)
              B.\(\dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{8}=1\)
              C.\(\dfrac{{{x}^{2}}}{9}+{{y}^{2}}=1\)
              D.\({{x}^{2}}+\dfrac{{{y}^{2}}}{9}=1\)
            • 3.
              方程\((x^{2}+y^{2}-2) \sqrt {x-3}=0\)表示的曲线是\((\)  \()\)
              A.一个圆和一条直线
              B.一个圆和一条射线
              C.一个圆
              D.一条直线
            • 4.
              设\(a > 0\)为常数,动点\(M(x,y)(y\neq 0)\)分别与两定点\(F_{1}(-a,0)\),\(F_{2}(a,0)\)的连线的斜率之积为定值\(λ\),若点\(M\)的轨迹是离心率为\( \sqrt {3}\)双曲线,则\(λ\)的值为\((\)  \()\)
              A.\(2\)
              B.\(-2\)
              C.\(3\)
              D.\( \sqrt {3}\)
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