What is 4x^2+4y^2+32x-24y+51=0 ?

The solution to 4x^2+4y^2+32x-24y+51=0 is Circle with (a,b)=(-4,3),r= 7/2

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4x^2+4y^2+32x-24y+51=0

4 x^{2} + 4 y^{2} + 32 x - 24 y + 51 = 0

(a, b) = (- 4, 3), r = \frac{7}{2}

Solution steps

4 x^{2} + 4 y^{2} + 32 x - 24 y + 51 = 0

Rewrite

4 x^{2} + 4 y^{2} + 32 x - 24 y + 51 = 0

in the form of the standard circle equation

(x - (- 4))^{2} + (y - 3)^{2} = (\frac{7}{2})^{2}

Therefore the circle properties are:

(a, b) = (- 4, 3), r = \frac{7}{2}

3 x^{2} - 12 x - 18 y + 3 y^{2} = 0

x^{2} + y^{2} - 10 x - 2 y + 22 = 0

(x + 3)^{2} + (y - 4)^{2} = 5^{2}

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

x^{2} + y^{2} - 7 x - 4 y = 0

What is 4x^2+4y^2+32x-24y+51=0 ?
The solution to 4x^2+4y^2+32x-24y+51=0 is Circle with (a,b)=(-4,3),r= 7/2