What is 3x^2+3y^2-21x+7=0 ?

The solution to 3x^2+3y^2-21x+7=0 is Circle with (a,b)=(7/2 ,0),r=(sqrt(119))/(2sqrt(3))

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3x^2+3y^2-21x+7=0

3 x^{2} + 3 y^{2} - 21 x + 7 = 0

(a, b) = (\frac{7}{2}, 0), r = \frac{119}{2 3}

Solution steps

3 x^{2} + 3 y^{2} - 21 x + 7 = 0

Rewrite

3 x^{2} + 3 y^{2} - 21 x + 7 = 0

in the form of the standard circle equation

(x - \frac{7}{2})^{2} + (y - 0)^{2} = (\frac{119}{2 3})^{2}

Therefore the circle properties are:

(a, b) = (\frac{7}{2}, 0), r = \frac{119}{2 3}

(x + \frac{19}{2})^{2} + (y - 230)^{2} = 40

x^{2} - 22 x + 49 = - y^{2} + 14 y

x^{2} - 2 x + y^{2} - 4 y = 11

x^{2} + y^{2} - 6 x + 4 y = 0

(y - 1)^{2} + (x - 5)^{2} = 1

What is 3x^2+3y^2-21x+7=0 ?
The solution to 3x^2+3y^2-21x+7=0 is Circle with (a,b)=(7/2 ,0),r=(sqrt(119))/(2sqrt(3))