Question 740852
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What is the center and radius of the circle x^2+y^2-8x-10y+32=0 ?
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x^2 + y^2 - 8x - 10y + 32 = 0,

x^2 + y^2 - 8x - 10y = -32.


Group the terms with x (one group) and y (another group)

(x^2 - 8x) + (y^2 - 10y) = -32.


Complete the squares

(x^2 - 8x + 16) + (y^2 - 10y + 25) = -32 + 16 + 25

(x-4)^2 + (y-5)^2 = 9.


It is the standard equation of the circle of the radius  {{{sqrt(9)}}} = 3 
with the center at (4,5).
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Solved.


The answer in the post by @lynnlo is incorrect.