What is the center (x-2)^2+(y-5)^2=49 ?

The center (x-2)^2+(y-5)^2=49 is (2,5)

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center (x-2)^2+(y-5)^2=49

center

(x - 2)^{2} + (y - 5)^{2} = 49

(2, 5)

Solution steps

(x - 2)^{2} + (y - 5)^{2} = 49

Rewrite

(x - 2)^{2} + (y - 5)^{2} = 49

in the form of the standard circle equation

(x - 2)^{2} + (y - 5)^{2} = 7^{2}

Therefore the circle properties are:

(a, b) = (2, 5), r = 7

And the center is:

(2, 5)

\frac{( x + 7 ) ^{2}}{16} + \frac{( y + 5 ) ^{2}}{16} = 1

x^{2} + 4 x + y^{2} + 6 y + 9 = 0

(x - 2)^{2} + (y + 6)^{2} = 25

(x - 1)^{2} + y^{2} = 4

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