What is the radius x^2+y^2+4x+14y+52=0 ?

The radius x^2+y^2+4x+14y+52=0 is r=1

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radius x^2+y^2+4x+14y+52=0

radius

x^{2} + y^{2} + 4 x + 14 y + 52 = 0

r = 1

Solution steps

x^{2} + y^{2} + 4 x + 14 y + 52 = 0

Rewrite

x^{2} + y^{2} + 4 x + 14 y + 52 = 0

in the form of the standard circle equation

(x - (- 2))^{2} + (y - (- 7))^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (- 2, - 7), r = 1

And the radius is:

r = 1

x^{2} + y^{2} - 14 x + 8 y - 16 = 0

(x - 3)^{2} + (y + 3)^{2} = 36

(x + 3)^{2} + (y + 1)^{2} = 4

x^{2} + y^{2} + 6 x + 2 y - 54 = 0

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