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

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

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

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

(a, b) = (3, - 4), r = 2

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (3, - 4), r = 2

2 x^{2} + 2 y^{2} + 10 x - 2 y = 7

(x - (- 5))^{2} + (y - 7)^{2} = (11)^{2}

x^{2} + y^{2} + 10 x - 8 y + 36 = 0

(x - 1)^{2} + (y + 11)^{2} + (0 - 8)^{2} = 100

x^{2} + y^{2} + 5 x - 2 y - 8 = 0

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