What is the radius (x+1)^2+(y+3)^2=5 ?

The radius (x+1)^2+(y+3)^2=5 is r=sqrt(5)

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radius (x+1)^2+(y+3)^2=5

radius

(x + 1)^{2} + (y + 3)^{2} = 5

r = 5

Solution steps

(x + 1)^{2} + (y + 3)^{2} = 5

Rewrite

(x + 1)^{2} + (y + 3)^{2} = 5

in the form of the standard circle equation

(x - (- 1))^{2} + (y - (- 3))^{2} = (5)^{2}

Therefore the circle properties are:

(a, b) = (- 1, - 3), r = 5

And the radius is:

r = 5

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

x^{2} + y^{2} + 7 x + 5 y + 16 = 0

(x + \frac{8}{3})^{2} + y^{2} = 1

x^{2} - 6 x + y^{2} + 8 y = 0

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

What is the radius (x+1)^2+(y+3)^2=5 ?
The radius (x+1)^2+(y+3)^2=5 is r=sqrt(5)