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已知圆\( C\)的方程\({x^2} + {y^2} - 2ax + \left( {2 - 4a} \right)y + 4a - 4 = 0\left( {a \in R} \right)\).
对\(a \in R\),求面积最小的圆\( C\)的方程.
参考答案:解:面积最小的圆就是以\(AB\)为一条直径的圆,圆心为\(\left( {\frac{4}{5},\frac{3}{5}} \right)\),半径为\(\frac{{3\sqrt 5 }}{5}\),方程是\({\left( {x - \frac{4}{5}} \right)^2} + {\left( {y - \frac{3}{5}} \right)^2} = \frac{9}{5}\).
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