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

The radius x^2+y^2+10x+4y+4=0 is r=5

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

radius

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

r = 5

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the radius is:

r = 5

3 x^{2} + 3 y^{2} - 2 x + y - 5 = 0

3 x^{2} - 12 x - 18 y + 3 y^{2} = 0

(x - 1)^{2} + (y + 4)^{2} = 9

x^{2} + y^{2} - 3 x + 4 y + 4 = 0

y^{2} = 4 - x^{2}