What is the center (x-12)^2+(y-7)^2=196 ?

The center (x-12)^2+(y-7)^2=196 is (12,7)

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center (x-12)^2+(y-7)^2=196

center

(x - 12)^{2} + (y - 7)^{2} = 196

(12, 7)

Solution steps

(x - 12)^{2} + (y - 7)^{2} = 196

Rewrite

(x - 12)^{2} + (y - 7)^{2} = 196

in the form of the standard circle equation

(x - 12)^{2} + (y - 7)^{2} = 1 4^{2}

Therefore the circle properties are:

(a, b) = (12, 7), r = 14

And the center is:

(12, 7)

9 x^{2} + 36 x + 9 y^{2} + 72 y + 155 = 0

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

\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

What is the center (x-12)^2+(y-7)^2=196 ?
The center (x-12)^2+(y-7)^2=196 is (12,7)