What is the radius 16x^2+16y^2-16x+24y-3=0 ?

The radius 16x^2+16y^2-16x+24y-3=0 is r=1

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radius 16x^2+16y^2-16x+24y-3=0

radius

16 x^{2} + 16 y^{2} - 16 x + 24 y - 3 = 0

r = 1

Solution steps

16 x^{2} + 16 y^{2} - 16 x + 24 y - 3 = 0

Rewrite

16 x^{2} + 16 y^{2} - 16 x + 24 y - 3 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, - \frac{3}{4}), r = 1

And the radius is:

r = 1

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

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

What is the radius 16x^2+16y^2-16x+24y-3=0 ?
The radius 16x^2+16y^2-16x+24y-3=0 is r=1