What is the center [x-7]^2+[y-3]^2=49 ?

The center [x-7]^2+[y-3]^2=49 is (7,3)

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center [x-7]^2+[y-3]^2=49

center

[x - 7]^{2} + [y - 3]^{2} = 49

(7, 3)

Solution steps

(x - 7)^{2} + (y - 3)^{2} = 49

Rewrite

(x - 7)^{2} + (y - 3)^{2} = 49

in the form of the standard circle equation

(x - 7)^{2} + (y - 3)^{2} = 7^{2}

Therefore the circle properties are:

(a, b) = (7, 3), r = 7

And the center is:

(7, 3)

x^{2} + 2 x + y^{2} = 0

2 x^{2} + 2 y^{2} + 8 x - 4 y - 15 = 0

x^{2} + y^{2} + 4 x + 14 y + 52 = 0

(x - 2)^{2} + (y - 4)^{2} = 49

x^{2} + y^{2} + 14 x - 2 y + 25 = 0