若圆C与圆\({\left( {x + 2} \right)^2} + {\left( {y - 1} \right)^2} = 1\)关于-可可试卷app

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若圆C与圆\({\left( {x + 2} \right)^2} + {\left( {y - 1} \right)^2} = 1\)关于原点对称，则圆C的方程为( ).

A.\({\left( {x - 2} \right)^2} + {\left( {y + 1} \right)^2} = 1\)

B.\({\left( {x - 2} \right)^2} + {\left( {y - 1} \right)^2} = 1\)

C.\({\left( {x - 1} \right)^2} + {\left( {y + 2} \right)^2} = 1\)

D.\({\left( {x + 1} \right)^2} + {\left( {y - 2} \right)^2} = 1\)

参考答案：A