What is 32=(x-3)^2+(y+1)^2 ?

The solution to 32=(x-3)^2+(y+1)^2 is Circle with (a,b)=(3,-1),r=4sqrt(2)

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32=(x-3)^2+(y+1)^2

32 = (x - 3)^{2} + (y + 1)^{2}

(a, b) = (3, - 1), r = 42

Solution steps

32 = (x - 3)^{2} + (y + 1)^{2}

Rewrite

32 = (x - 3)^{2} + (y + 1)^{2}

in the form of the standard circle equation

(x - 3)^{2} + (y - (- 1))^{2} = (42)^{2}

Therefore the circle properties are:

(a, b) = (3, - 1), r = 42

x^{2} - 10 x + y^{2} + 6 y - 30 = 0

x^{2} - 2 x + 1 + y^{2} + y + \frac{1}{11} = \frac{11}{4}

(- \frac{2}{5} - x)^{2} + (- \frac{6}{5} - y)^{2} = 10

9 x^{2} + 9 y^{2} - 24 x + 84 y + 65 = 0

(x - 1.5)^{2} + (y - 1)^{2} = 3

What is 32=(x-3)^2+(y+1)^2 ?
The solution to 32=(x-3)^2+(y+1)^2 is Circle with (a,b)=(3,-1),r=4sqrt(2)