What is the radius x^2+y^2-10x+4y+25=0 ?

The radius x^2+y^2-10x+4y+25=0 is r=2

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radius x^2+y^2-10x+4y+25=0

radius

x^{2} + y^{2} - 10 x + 4 y + 25 = 0

r = 2

Solution steps

x^{2} + y^{2} - 10 x + 4 y + 25 = 0

Rewrite

x^{2} + y^{2} - 10 x + 4 y + 25 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the radius is:

r = 2

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

x^{2} + y^{2} - 64 = 0

3 x^{2} + 5 x + 3 y^{2} + 6 y - 15 = 0

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

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