What is the radius 4x^2+4y^2-8y-5=0 ?

The radius 4x^2+4y^2-8y-5=0 is r= 3/2

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radius 4x^2+4y^2-8y-5=0

radius

4 x^{2} + 4 y^{2} - 8 y - 5 = 0

r = \frac{3}{2}

Solution steps

4 x^{2} + 4 y^{2} - 8 y - 5 = 0

Rewrite

4 x^{2} + 4 y^{2} - 8 y - 5 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - 1)^{2} = (\frac{3}{2})^{2}

Therefore the circle properties are:

(a, b) = (0, 1), r = \frac{3}{2}

And the radius is:

r = \frac{3}{2}

x^{2} + y^{2} - 7 x = 36 - 4 y

(x - 5)^{2} + (y + 1)^{2} = 16

(x - 3)^{2} + (y + 4)^{2} = 16

x^{2} + y^{2} + 4 x + 20 y + 23 = 0

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