x²+y²–4x+2my+2m²–2m+1=0(x-2)²-4+(y+m)²+m²-2m+1=0(x-2)²+(y+m)²=-m²+2m+3所以-m²+2m+3＞0即m²-2m-3＜0(m-3)(m+1)＜0解得-1＜m＜3即当-1＜m＜3时，方程表示圆-m²+2m+3=-(m²-2m)+3=-(m-1)²+4当m=1时有最大值为4即半径最大时的方程为(x-2)²+(y+1)²=4
