What is 4x^2+4y^2+16x-12y+1=0 ?

The solution to 4x^2+4y^2+16x-12y+1=0 is Circle with (a,b)=(-2, 3/2),r=sqrt(6)

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4x^2+4y^2+16x-12y+1=0

4 x^{2} + 4 y^{2} + 16 x - 12 y + 1 = 0

(a, b) = (- 2, \frac{3}{2}), r = 6

Solution steps

4 x^{2} + 4 y^{2} + 16 x - 12 y + 1 = 0

Rewrite

4 x^{2} + 4 y^{2} + 16 x - 12 y + 1 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 2, \frac{3}{2}), r = 6

x^{2} + y^{2} + 84 x - 22 y + 861 = 0

x^{2} + y^{2} + 6 x - 6 y + 9 = 0

x^{2} + y^{2} = (53)^{2}

3 (x - 5)^{2} + 3 y^{2} = 27

(x - 4)^{2} + y^{2} = 9

What is 4x^2+4y^2+16x-12y+1=0 ?
The solution to 4x^2+4y^2+16x-12y+1=0 is Circle with (a,b)=(-2, 3/2),r=sqrt(6)