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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + 14 x + y^{2} + 10 y = - 62

(x + 6)^{2} + (y - 2)^{2} = 4

(x - 5)^{2} + (y - 1)^{2} = 49

(x - 1)^{2} + (y + 5)^{2} = 25

x^{2} + y^{2} - \frac{12}{11} x - \frac{84}{11} y - \frac{640}{11} = 0

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