What is 4x^2+4y^2+6x-8y-4=0 ?

The solution to 4x^2+4y^2+6x-8y-4=0 is Circle with (a,b)=(-3/4 ,1),r=(sqrt(41))/4

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4x^2+4y^2+6x-8y-4=0

4 x^{2} + 4 y^{2} + 6 x - 8 y - 4 = 0

(a, b) = (- \frac{3}{4}, 1), r = \frac{41}{4}

Solution steps

4 x^{2} + 4 y^{2} + 6 x - 8 y - 4 = 0

Rewrite

4 x^{2} + 4 y^{2} + 6 x - 8 y - 4 = 0

in the form of the standard circle equation

(x - (- \frac{3}{4}))^{2} + (y - 1)^{2} = (\frac{41}{4})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{3}{4}, 1), r = \frac{41}{4}

x^{2} + y^{2} - 10 x - 12 y - 3 = 0

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

x^{2} + y^{2} = \frac{1}{4}

x^{2} + 8 x + y^{2} + 15 = 0

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

What is 4x^2+4y^2+6x-8y-4=0 ?
The solution to 4x^2+4y^2+6x-8y-4=0 is Circle with (a,b)=(-3/4 ,1),r=(sqrt(41))/4