What is x^2+4x+2^2+y^2+8y+4^2-5=20 ?

The solution to x^2+4x+2^2+y^2+8y+4^2-5=20 is Circle with (a,b)=(-2,-4),r=5

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x^2+4x+2^2+y^2+8y+4^2-5=20

x^{2} + 4 x + 2^{2} + y^{2} + 8 y + 4^{2} - 5 = 20

(a, b) = (- 2, - 4), r = 5

Solution steps

x^{2} + 4 x + 2^{2} + y^{2} + 8 y + 4^{2} - 5 = 20

Rewrite

x^{2} + 4 x + 2^{2} + y^{2} + 8 y + 4^{2} - 5 = 20

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 2, - 4), r = 5

x^{2} + y^{2} - 5 x - 3 y = 0

9 x^{2} + 72 x + 9 y^{2} - 12 y + 103 = 0

(x + 3)^{2} + (y - 2)^{2} = 64

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

x^{2} + y^{2} - 10 x - 14 y - 7 = 0

What is x^2+4x+2^2+y^2+8y+4^2-5=20 ?
The solution to x^2+4x+2^2+y^2+8y+4^2-5=20 is Circle with (a,b)=(-2,-4),r=5