(x−7)2+(y−4)2y=8x−x2−12y2=−(x−4)2+16−12(x−4)2+y2=4 
m=9M=41M2−m2=412−92=1600
Let the maximum and minimum values of (8x−x2−12−4)2+(x−7)2,x∈R be M and m, respectively. Then M2−m2 is equal to _________
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