What is (x-1)^2+(y+3)^2=2 ?

The solution to (x-1)^2+(y+3)^2=2 is Circle with (a,b)=(1,-3),r=sqrt(2)

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(x-1)^2+(y+3)^2=2

(x - 1)^{2} + (y + 3)^{2} = 2

(a, b) = (1, - 3), r = 2

Solution steps

(x - 1)^{2} + (y + 3)^{2} = 2

Rewrite

(x - 1)^{2} + (y + 3)^{2} = 2

in the form of the standard circle equation

(x - 1)^{2} + (y - (- 3))^{2} = (2)^{2}

Therefore the circle properties are:

(a, b) = (1, - 3), r = 2

(x + 3)^{2} + (y + 2)^{2} = 25

36 = \frac{1}{49} (x^{2} + y^{2})

4 x^{2} + 4 y^{2} - 24 y - 13 = 0

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

x^{2} + 4 x + y^{2} - 4 y = 20

What is (x-1)^2+(y+3)^2=2 ?
The solution to (x-1)^2+(y+3)^2=2 is Circle with (a,b)=(1,-3),r=sqrt(2)