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已知圆\({C_1}:{\left( {x - a} \right)^2} + {\left( {y + 2} \right)^2} = 4\)与圆\({C_2}:{\left( {x + b} \right)^2} + {\left( {y + 2} \right)^2} = 1\)相内切， 则\({a^2} + {b^2}\)的最小值为___．
参考答案：\(\frac {1} {2}\)
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