What is 3x^2-2x+3+3y^2+8y+48= 172/3 ?

The solution to 3x^2-2x+3+3y^2+8y+48= 172/3 is Circle with (a,b)=(1/3 ,-4/3),r=2

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3x^2-2x+3+3y^2+8y+48= 172/3

3 x^{2} - 2 x + 3 + 3 y^{2} + 8 y + 48 = \frac{172}{3}

(a, b) = (\frac{1}{3}, - \frac{4}{3}), r = 2

Solution steps

3 x^{2} - 2 x + 3 + 3 y^{2} + 8 y + 48 = \frac{172}{3}

Rewrite

3 x^{2} - 2 x + 3 + 3 y^{2} + 8 y + 48 = \frac{172}{3}

in the form of the standard circle equation

(x - \frac{1}{3})^{2} + (y - (- \frac{4}{3}))^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{3}, - \frac{4}{3}), r = 2

(x - 1)^{2} + (y + 5)^{2} < 4

(X + 4)^{2} + (Y - 1)^{2} = 148

(x + 2)^{2} + (y + 3)^{2} = 11

x^{2} + y^{2} + 4 x - 4 y = 1

(x + 2)^{2} + (y + 3)^{2} = 15

What is 3x^2-2x+3+3y^2+8y+48= 172/3 ?
The solution to 3x^2-2x+3+3y^2+8y+48= 172/3 is Circle with (a,b)=(1/3 ,-4/3),r=2