What is 4x^2+4y^2-8x+16y+4=0 ?

The solution to 4x^2+4y^2-8x+16y+4=0 is Circle with (a,b)=(1,-2),r=2

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4x^2+4y^2-8x+16y+4=0

4 x^{2} + 4 y^{2} - 8 x + 16 y + 4 = 0

(a, b) = (1, - 2), r = 2

Solution steps

4 x^{2} + 4 y^{2} - 8 x + 16 y + 4 = 0

Rewrite

4 x^{2} + 4 y^{2} - 8 x + 16 y + 4 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (1, - 2), r = 2

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

x^{2} + y^{2} + 2 (\frac{29}{22}) x + 2 (\frac{29}{22}) y + (- \frac{142}{11}) = 0

(x - 7)^{2} + (y + 3)^{2} - 81 = 0

3 x^{2} + 3 y^{2} = 9

3 x^{2} + 3 y^{2} = 6

What is 4x^2+4y^2-8x+16y+4=0 ?
The solution to 4x^2+4y^2-8x+16y+4=0 is Circle with (a,b)=(1,-2),r=2