It is known that P (x, y) is a circle x ^ 2 + y ^ 2-6x-4y + 12 = 0 Then x ^ 2 + y ^ 2 value range The answer is 14 + or 13 under - 2 radical | Math enthusiast | Mathematical art

It is known that P (x, y) is a circle x ^ 2 + y ^ 2-6x-4y + 12 = 0 Then x ^ 2 + y ^ 2 value range The answer is 14 + or 13 under - 2 radical

x^2+y^2-6x-4y+12=0
（x-3）^2+(y-2)^2=1
The center of the circle is (3,2) and the radius is 1
So, the range is the sum of 3 squares plus 2 squares plus / minus 1
It's (root 13-1, root 13 + 1)

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