What is x^2+(y-25/3)^2=144 ?

The solution to x^2+(y-25/3)^2=144 is Circle with (a,b)=(0, 25/3),r=12

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x^2+(y-25/3)^2=144

x^{2} + (y - \frac{25}{3})^{2} = 144

(a, b) = (0, \frac{25}{3}), r = 12

Solution steps

x^{2} + (y - \frac{25}{3})^{2} = 144

Rewrite

x^{2} + (y - \frac{25}{3})^{2} = 144

in the form of the standard circle equation

(x - 0)^{2} + (y - \frac{25}{3})^{2} = 1 2^{2}

Therefore the circle properties are:

(a, b) = (0, \frac{25}{3}), r = 12

\frac{x ^{2}}{15} + \frac{y ^{2}}{15} - \frac{2 x}{15} - \frac{6}{15} = 0

(x - 7)^{2} + (y + 13)^{2} = 81

x^{2} - 7 x + y^{2} - y - 2 = 0

(x - 8)^{2} + (y - 3)^{2} = 3

(x^{2} + 5 x + \frac{25}{4}) + y^{2} = \frac{25}{4} + \frac{25}{4}

What is x^2+(y-25/3)^2=144 ?
The solution to x^2+(y-25/3)^2=144 is Circle with (a,b)=(0, 25/3),r=12