What is (x+2)^2+(y+sqrt(5))^2=7 ?

The solution to (x+2)^2+(y+sqrt(5))^2=7 is Circle with (a,b)=(-2,-sqrt(5)),r=sqrt(7)

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(x+2)^2+(y+sqrt(5))^2=7

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

(a, b) = (- 2, - 5), r = 7

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 2, - 5), r = 7

4 x^{2} + 4 y^{2} - 4 x + 24 y + 1 = 0

(x - 4)^{2} + (y - 7)^{2} = 49

9 x^{2} + 9 y^{2} - 42 x + 84 y + 65 = 0

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

2 x^{2} + 8 x + 2 y^{2} - 16 y - 5 = 0

What is (x+2)^2+(y+sqrt(5))^2=7 ?
The solution to (x+2)^2+(y+sqrt(5))^2=7 is Circle with (a,b)=(-2,-sqrt(5)),r=sqrt(7)