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

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

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

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

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

Solution steps

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

Rewrite

4 x^{2} + 40 x + 4 y^{2} - 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}

x^{2} + y^{2} - 6 x - 2 y + 1 = 0

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

x^{2} + y^{2} + 4 x - 12 y + 3 = 0

x^{2} + y^{2} - 6 x + 5 = 2

(x - 1)^{2} + (y - 4)^{2} = 25

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