What is 4x^2+4y^2-32y+39=0 ?

The solution to 4x^2+4y^2-32y+39=0 is Circle with (a,b)=(0,4),r= 5/2

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4x^2+4y^2-32y+39=0

4 x^{2} + 4 y^{2} - 32 y + 39 = 0

(a, b) = (0, 4), r = \frac{5}{2}

Solution steps

4 x^{2} + 4 y^{2} - 32 y + 39 = 0

Rewrite

4 x^{2} + 4 y^{2} - 32 y + 39 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - 4)^{2} = (\frac{5}{2})^{2}

Therefore the circle properties are:

(a, b) = (0, 4), r = \frac{5}{2}

x^{2} + 12 x + y^{2} - 12 y = - 56

(x - 1)^{2} + (y + 6)^{2} \leq 4

(x - 10)^{2} + (y + 1)^{2} - 64 = 0

- 3 y^{2} = 3 x^{2} - 2 x - 1

45 = (x - 1)^{2} + (y - (- 3))^{2}

What is 4x^2+4y^2-32y+39=0 ?
The solution to 4x^2+4y^2-32y+39=0 is Circle with (a,b)=(0,4),r= 5/2