What is 4x^2+4y^2+40x+32y=5 ?

The solution to 4x^2+4y^2+40x+32y=5 is Circle with (a,b)=(-5,-4),r= 13/2

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4x^2+4y^2+40x+32y=5

4 x^{2} + 4 y^{2} + 40 x + 32 y = 5

(a, b) = (- 5, - 4), r = \frac{13}{2}

Solution steps

4 x^{2} + 4 y^{2} + 40 x + 32 y = 5

Rewrite

4 x^{2} + 4 y^{2} + 40 x + 32 y = 5

in the form of the standard circle equation

(x - (- 5))^{2} + (y - (- 4))^{2} = (\frac{13}{2})^{2}

Therefore the circle properties are:

(a, b) = (- 5, - 4), r = \frac{13}{2}

9 x^{2} + 9 y^{2} - 6 x - 12 y - 11 = 0

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

x^{2} + y^{2} - 26 x + 165 = 0

(x + 4)^{2} + (y + 6)^{2} = 16

(x - 8)^{2} + (y + 2)^{2} = 4

What is 4x^2+4y^2+40x+32y=5 ?
The solution to 4x^2+4y^2+40x+32y=5 is Circle with (a,b)=(-5,-4),r= 13/2