What is the radius x^2+y^2+2x+4y-20=0 ?

The radius x^2+y^2+2x+4y-20=0 is r=5

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radius x^2+y^2+2x+4y-20=0

radius

x^{2} + y^{2} + 2 x + 4 y - 20 = 0

r = 5

Solution steps

x^{2} + y^{2} + 2 x + 4 y - 20 = 0

Rewrite

x^{2} + y^{2} + 2 x + 4 y - 20 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 1, - 2), r = 5

And the radius is:

r = 5

(x - 2)^{2} + (y - 5)^{2} = 9

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

(x - 2)^{2} + (y + 3)^{2} = 9

3 y^{2} + 3 x^{2} + 24 x + 32 = 0

(x + 4)^{2} + (y - 3)^{2} = 25