What is (x-10)^2+(y-5)^2=28 ?

The solution to (x-10)^2+(y-5)^2=28 is Circle with (a,b)=(10,5),r=sqrt(28)

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(x-10)^2+(y-5)^2=28

(x - 10)^{2} + (y - 5)^{2} = 28

(a, b) = (10, 5), r = 28

Solution steps

(x - 10)^{2} + (y - 5)^{2} = 28

Rewrite

(x - 10)^{2} + (y - 5)^{2} = 28

in the form of the standard circle equation

(x - 10)^{2} + (y - 5)^{2} = (28)^{2}

Therefore the circle properties are:

(a, b) = (10, 5), r = 28

f (x) (f (x)) = - x^{2} + 3 x

x^{2} - 6 x + y^{2} - 4 y = - 4

x^{2} + y^{2} - 18 x + 12 y + 80 = 0

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

x^{2} + y^{2} - 280 x + 4244 y + 464 = 0

What is (x-10)^2+(y-5)^2=28 ?
The solution to (x-10)^2+(y-5)^2=28 is Circle with (a,b)=(10,5),r=sqrt(28)