What is (x-3)^2+(y-4)^2-9=0 ?

The solution to (x-3)^2+(y-4)^2-9=0 is Circle with (a,b)=(3,4),r=3

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(x-3)^2+(y-4)^2-9=0

(x - 3)^{2} + (y - 4)^{2} - 9 = 0

(a, b) = (3, 4), r = 3

Solution steps

(x - 3)^{2} + (y - 4)^{2} - 9 = 0

Rewrite

(x - 3)^{2} + (y - 4)^{2} - 9 = 0

in the form of the standard circle equation

(x - 3)^{2} + (y - 4)^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (3, 4), r = 3

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

3^{2} = (x - 2)^{2} + (y - 4)^{2}

(x - 5.5)^{2} + (y - 5.5)^{2} - (1.5)^{2} = 0

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

(x - 3)^{2} + (y - 6)^{2} = 2

What is (x-3)^2+(y-4)^2-9=0 ?
The solution to (x-3)^2+(y-4)^2-9=0 is Circle with (a,b)=(3,4),r=3