What is the radius x^2+y^2-4x+10y+20=0 ?

The radius x^2+y^2-4x+10y+20=0 is r=3

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radius x^2+y^2-4x+10y+20=0

radius

x^{2} + y^{2} - 4 x + 10 y + 20 = 0

r = 3

Solution steps

x^{2} + y^{2} - 4 x + 10 y + 20 = 0

Rewrite

x^{2} + y^{2} - 4 x + 10 y + 20 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the radius is:

r = 3

4 x^{2} + 4 y^{2} + 40 x - 32 y = 5

(x - 3)^{2} + (y + 8)^{2} = 35

(y + 8)^{2} + (x + 2)^{2} = 72

5 x^{2} - 20 x + 5 y^{2} + 30 y - 11 = 0

x^{2} + y^{2} - 6 x + 10 y + 9 = 0