What is (x+2)^2+(y+3)^2=5 ?

The solution to (x+2)^2+(y+3)^2=5 is Circle with (a,b)=(-2,-3),r=sqrt(5)

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(x+2)^2+(y+3)^2=5

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

(a, b) = (- 2, - 3), r = 5

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 2, - 3), r = 5

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

(x - 5)^{2} + (y + 6)^{2} = 25

x^{2} + y^{2} + 6 x + 2 y - 26 = 0

(x - 9)^{2} + (y + 7)^{2} = 169

x^{2} + y^{2} \leq 5

What is (x+2)^2+(y+3)^2=5 ?
The solution to (x+2)^2+(y+3)^2=5 is Circle with (a,b)=(-2,-3),r=sqrt(5)