What is 4x^2+4y^2+40x+16y+40=0 ?

The solution to 4x^2+4y^2+40x+16y+40=0 is Circle with (a,b)=(-5,-2),r=sqrt(19)

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4x^2+4y^2+40x+16y+40=0

4 x^{2} + 4 y^{2} + 40 x + 16 y + 40 = 0

(a, b) = (- 5, - 2), r = 19

Solution steps

4 x^{2} + 4 y^{2} + 40 x + 16 y + 40 = 0

Rewrite

4 x^{2} + 4 y^{2} + 40 x + 16 y + 40 = 0

in the form of the standard circle equation

(x - (- 5))^{2} + (y - (- 2))^{2} = (19)^{2}

Therefore the circle properties are:

(a, b) = (- 5, - 2), r = 19

(x + 1)^{2} + (y + 3)^{2} = 70. 9^{2}

(x - 21)^{2} + (y - 2)^{2} = 16

x^{2} - 2 x + y^{2} - 8 y - 20 = 0

y^{2} = - x^{2} + 10

x^{2} + y^{2} = 3600

What is 4x^2+4y^2+40x+16y+40=0 ?
The solution to 4x^2+4y^2+40x+16y+40=0 is Circle with (a,b)=(-5,-2),r=sqrt(19)