What is the radius x^2+y^2-8*+7=0 ?

The radius x^2+y^2-8*+7=0 is r=1

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radius x^2+y^2-8*+7=0

radius

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

r = 1

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 0)^{2} + (y - 0)^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (0, 0), r = 1

And the radius is:

r = 1

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

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

y^{2} + x^{2} = 25

x^{2} + y^{2} - 14 x - 12 y + 59 = 0

(x + 5)^{2} + (y - 2)^{2} = 25