What is 4x^2-4x+4y^2-35=0 ?

The solution to 4x^2-4x+4y^2-35=0 is Circle with (a,b)=(1/2 ,0),r=3

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4x^2-4x+4y^2-35=0

4 x^{2} - 4 x + 4 y^{2} - 35 = 0

(a, b) = (\frac{1}{2}, 0), r = 3

Solution steps

4 x^{2} - 4 x + 4 y^{2} - 35 = 0

Rewrite

4 x^{2} - 4 x + 4 y^{2} - 35 = 0

in the form of the standard circle equation

(x - \frac{1}{2})^{2} + (y - 0)^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, 0), r = 3

y^{2} = 10 - z^{2}

3 x^{2} + 3 y^{2} - 6 x + 5 y = 0

x^{2} - 8 x + y^{2} - 20 = 0

16 x^{2} + 16 y^{2} + 16 x + 40 y - 7 = 0

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

What is 4x^2-4x+4y^2-35=0 ?
The solution to 4x^2-4x+4y^2-35=0 is Circle with (a,b)=(1/2 ,0),r=3