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

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

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

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

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

Solution steps

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

Rewrite

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

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

(x - 9)^{2} + (y + 1)^{2} = 36

9 x^{2} + 9 y^{2} = 4

x^{2} + y^{2} = \frac{4}{9}

(x - 3)^{2} + (y - 8)^{2} = 5^{2}

x^{2} + y^{2} + 2 x - 8 y + 13 = 0

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