What is the radius x^2+y^2+8x-14y-56=0 ?

The radius x^2+y^2+8x-14y-56=0 is r=11

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radius x^2+y^2+8x-14y-56=0

radius

x^{2} + y^{2} + 8 x - 14 y - 56 = 0

r = 11

Solution steps

x^{2} + y^{2} + 8 x - 14 y - 56 = 0

Rewrite

x^{2} + y^{2} + 8 x - 14 y - 56 = 0

in the form of the standard circle equation

(x - (- 4))^{2} + (y - 7)^{2} = 1 1^{2}

Therefore the circle properties are:

(a, b) = (- 4, 7), r = 11

And the radius is:

r = 11

x^{2} + y^{2} + 3 x - 5 y + 1 = 0

x^{2} + y^{2} = 1 + 2 y

x^{2} + y^{2} - 8 \cdot + 7 = 0

(x + 2)^{2} + (y + 5)^{2} = 16

x^{2} + y^{2} + 7 x - 4 y - 2 = 0