What is the center (x-3)^2+(y-5)^2=4 ?

The center (x-3)^2+(y-5)^2=4 is (3,5)

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center (x-3)^2+(y-5)^2=4

center

(x - 3)^{2} + (y - 5)^{2} = 4

(3, 5)

Solution steps

(x - 3)^{2} + (y - 5)^{2} = 4

Rewrite

(x - 3)^{2} + (y - 5)^{2} = 4

in the form of the standard circle equation

(x - 3)^{2} + (y - 5)^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (3, 5), r = 2

And the center is:

(3, 5)

x^{2} + 20 x + y^{2} + 16 y = - 20

100 x^{2} + 100 y^{2} - 100 x + 400 y + 409 = 0

\frac{x ^{2}}{9} + \frac{y ^{2}}{9} = 1

(x + 3)^{2} + (y - 4)^{2} = 25

(x + 3)^{2} + (y - 4)^{2} = 16