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}

x^{2} + y^{2} + 8 y = 33

x^{2} + 2 x + y^{2} - 35 = 0

x^{2} + y^{2} + 12 x - 8 y + 3 = 0

8 x^{2} + 8 y^{2} - 20 x + 64 y - 371 = 0

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

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