已知圆C1:x^2+y^2=4与圆C2:x^2+y^2+2ax+2by=0,(a^2+b^2≠0)的公共弦长为1,求动圆C2的圆心C2所在曲线的方程
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![已知圆C1:x^2+y^2=4与圆C2:x^2+y^2+2ax+2by=0,(a^2+b^2≠0)的公共弦长为1,求动圆C2的圆心C2所在曲线的方程](/uploads/image/z/3772712-56-2.jpg?t=%E5%B7%B2%E7%9F%A5%E5%9C%86C1%3Ax%5E2%2By%5E2%3D4%E4%B8%8E%E5%9C%86C2%3Ax%5E2%2By%5E2%2B2ax%2B2by%3D0%2C%EF%BC%88a%5E2%2Bb%5E2%E2%89%A00%EF%BC%89%E7%9A%84%E5%85%AC%E5%85%B1%E5%BC%A6%E9%95%BF%E4%B8%BA1%2C%E6%B1%82%E5%8A%A8%E5%9C%86C2%E7%9A%84%E5%9C%86%E5%BF%83C2%E6%89%80%E5%9C%A8%E6%9B%B2%E7%BA%BF%E7%9A%84%E6%96%B9%E7%A8%8B)
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