What is 4x^2+4y^2-24x+8y+39=0 ?

The solution to 4x^2+4y^2-24x+8y+39=0 is Circle with (a,b)=(3,-1),r= 1/2

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4x^2+4y^2-24x+8y+39=0

4 x^{2} + 4 y^{2} - 24 x + 8 y + 39 = 0

(a, b) = (3, - 1), r = \frac{1}{2}

Solution steps

4 x^{2} + 4 y^{2} - 24 x + 8 y + 39 = 0

Rewrite

4 x^{2} + 4 y^{2} - 24 x + 8 y + 39 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (3, - 1), r = \frac{1}{2}

(x - 4)^{2} + (y - 5)^{2} = 64

(x)^{2} + (y)^{2} - 6 x + 10 y + 34 = 72

a - b + a^{2} + b^{2} = 0

2 x^{2} - 4 x + 2 y^{2} + 18 y = 7

(x - 1)^{2} + y^{2} - 1 = \frac{3}{6} x

What is 4x^2+4y^2-24x+8y+39=0 ?
The solution to 4x^2+4y^2-24x+8y+39=0 is Circle with (a,b)=(3,-1),r= 1/2