What is x^2+(y-3)^2-49=0 ?

The solution to x^2+(y-3)^2-49=0 is Circle with (a,b)=(0,3),r=7

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x^2+(y-3)^2-49=0

x^{2} + (y - 3)^{2} - 49 = 0

(a, b) = (0, 3), r = 7

Solution steps

x^{2} + (y - 3)^{2} - 49 = 0

Rewrite

x^{2} + (y - 3)^{2} - 49 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - 3)^{2} = 7^{2}

Therefore the circle properties are:

(a, b) = (0, 3), r = 7

2 x^{2} + 2 y^{2} - 6 x - 14 y - 3 = 0

4 x^{2} + 4 y^{2} - 16 x + 12 y + 21 = 0

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

(x + (- 3))^{2} + (y - 2)^{2} = 16

x^{2} + y^{2} + 8 x + 2 y + 1 = 0

What is x^2+(y-3)^2-49=0 ?
The solution to x^2+(y-3)^2-49=0 is Circle with (a,b)=(0,3),r=7