13. 在平面内,A(3,0)A(3,0)A(3,0),B(−1,0)B(-1,0)B(−1,0),CCC为动点,若AC→⋅BC→=5\overrightarrow{AC}\cdot \overrightarrow{BC}=5AC⋅BC=5.(1)求点CCC的轨迹方程;(2)若直线l:x−y+3=0l:x-y+3=0l:x−y+3=0与曲线CCC交于MMM,NNN,求∣MN∣\vert MN\vert∣MN∣的长.