What is 4x^{(2)}+4y^{(2)}-24x+16y+51=0 ?

The solution to 4x^{(2)}+4y^{(2)}-24x+16y+51=0 is Circle with (a,b)=(3,-2),r= 1/2

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4x^{(2)}+4y^{(2)}-24x+16y+51=0

4 x^{(2)} + 4 y^{(2)} - 24 x + 16 y + 51 = 0

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

Solution steps

4 x^{2} + 4 y^{2} - 24 x + 16 y + 51 = 0

Rewrite

4 x^{2} + 4 y^{2} - 24 x + 16 y + 51 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + y^{2} + 12 x - 10 y = - 45

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

x^{2} - 16 x + 64 + y^{2} - 14 x + 49 = 0

5 x^{2} - 5 x + 5 y^{2} - 4 y - 9 = 0

x^{2} - 2 x + y^{2} = 15

What is 4x^{(2)}+4y^{(2)}-24x+16y+51=0 ?
The solution to 4x^{(2)}+4y^{(2)}-24x+16y+51=0 is Circle with (a,b)=(3,-2),r= 1/2