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

The solution to 4(x+5)^2+4(y+4)^2-20=0 is Circle with (a,b)=(-5,-4),r=sqrt(5)

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

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

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

x^{2} + y^{2} - 2 x + 4 y - 21 = 0

(x - 1)^{2} + (y - 9)^{2} = 15

(x - 3)^{2} + (y - 4)^{2} = 7

169 - x^{2} - y^{2} > 0

(x + 8)^{2} + (y + 5)^{2} = 9

What is 4(x+5)^2+4(y+4)^2-20=0 ?
The solution to 4(x+5)^2+4(y+4)^2-20=0 is Circle with (a,b)=(-5,-4),r=sqrt(5)