What is x^2+(y-4)^2=1 ?

The solution to x^2+(y-4)^2=1 is Circle with (a,b)=(0,4),r=1

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

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

(a, b) = (0, 4), r = 1

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (0, 4), r = 1

y^{2} = 16 - x^{2}

x^{2} + y^{2} - 8 = 0

25 x^{2} + 25 y^{2} + 30 x - 20 y - 62 = 0

x^{2} + y^{2} - 81 = 0

(x - 3)^{2} + (y + 2)^{2} = 9

What is x^2+(y-4)^2=1 ?
The solution to x^2+(y-4)^2=1 is Circle with (a,b)=(0,4),r=1