What is 4y^2+4x^2-4y-4x-7=0 ?

The solution to 4y^2+4x^2-4y-4x-7=0 is Circle with (a,b)=(1/2 , 1/2),r= 3/2

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

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

x^{2} - 4 x + y^{2} + 5 y = 12

(x - 5)^{2} + (y + 3)^{2} = 1

2 (x - 5)^{2} + 2 y^{2} = 8

(x + 15)^{2} + (y + 11)^{2} = 25

What is 4y^2+4x^2-4y-4x-7=0 ?
The solution to 4y^2+4x^2-4y-4x-7=0 is Circle with (a,b)=(1/2 , 1/2),r= 3/2