What is (x-33/4)^2+y^2=25 ?

The solution to (x-33/4)^2+y^2=25 is Circle with (a,b)=(33/4 ,0),r=5

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(x-33/4)^2+y^2=25

(x - \frac{33}{4})^{2} + y^{2} = 25

(a, b) = (\frac{33}{4}, 0), r = 5

Solution steps

(x - \frac{33}{4})^{2} + y^{2} = 25

Rewrite

(x - \frac{33}{4})^{2} + y^{2} = 25

in the form of the standard circle equation

(x - \frac{33}{4})^{2} + (y - 0)^{2} = 5^{2}

Therefore the circle properties are:

(a, b) = (\frac{33}{4}, 0), r = 5

0 > (x - 1)^{2} + y^{2} - 1

x^{(2)} + y^{(2)} + 4 x - 4 y - 1 = 0

x, x^{2} = 8 - r (x)^{2}

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

x^{2} + y^{2} - 10 x + 4 y - 20 = 0

What is (x-33/4)^2+y^2=25 ?
The solution to (x-33/4)^2+y^2=25 is Circle with (a,b)=(33/4 ,0),r=5