What is 4x^2-12x+4y^2+16y-5=0 ?

The solution to 4x^2-12x+4y^2+16y-5=0 is Circle with (a,b)=(3/2 ,-2),r=sqrt(15/2)

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4x^2-12x+4y^2+16y-5=0

4 x^{2} - 12 x + 4 y^{2} + 16 y - 5 = 0

(a, b) = (\frac{3}{2}, - 2), r = \frac{15}{2}

Solution steps

4 x^{2} - 12 x + 4 y^{2} + 16 y - 5 = 0

Rewrite

4 x^{2} - 12 x + 4 y^{2} + 16 y - 5 = 0

in the form of the standard circle equation

(x - \frac{3}{2})^{2} + (y - (- 2))^{2} = (\frac{15}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{3}{2}, - 2), r = \frac{15}{2}

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

x^{2} + y^{2} - 72 = 0

x^{2} - 6 x + y^{2} - 4 y - 12 = 0

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

x^{2} + y^{2} + 28 x + 52 = 0

What is 4x^2-12x+4y^2+16y-5=0 ?
The solution to 4x^2-12x+4y^2+16y-5=0 is Circle with (a,b)=(3/2 ,-2),r=sqrt(15/2)