What is the radius 4(x+5)^2+4(y+4)^2-20=0 ?

The radius 4(x+5)^2+4(y+4)^2-20=0 is r=sqrt(5)

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radius 4(x+5)^2+4(y+4)^2-20=0

radius

4 (x + 5)^{2} + 4 (y + 4)^{2} - 20 = 0

r = 5

Solution steps

4 (x + 5)^{2} + 4 (y + 4)^{2} - 20 = 0

Rewrite

4 (x + 5)^{2} + 4 (y + 4)^{2} - 20 = 0

in the form of the standard circle equation

(x - (- 5))^{2} + (y - (- 4))^{2} = (5)^{2}

Therefore the circle properties are:

(a, b) = (- 5, - 4), r = 5

And the radius is:

r = 5

x^{2} + y^{2} + 2 x - 8 y - 19 = 0

x^{2} + y^{2} - 8 x - 13 y - 32 = 0

x^{2} + y^{2} - 6 x - 8 y + 9 = 0

(x - 9)^{2} + y^{2} - 32 = 0

(x - 3)^{2} = 8 - (y + 3)^{2}

What is the radius 4(x+5)^2+4(y+4)^2-20=0 ?
The radius 4(x+5)^2+4(y+4)^2-20=0 is r=sqrt(5)