What is 4x^2+4y^2+24x+16y+27=0 ?

The solution to 4x^2+4y^2+24x+16y+27=0 is Circle with (a,b)=(-3,-2),r= 5/2

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4x^2+4y^2+24x+16y+27=0

4 x^{2} + 4 y^{2} + 24 x + 16 y + 27 = 0

(a, b) = (- 3, - 2), r = \frac{5}{2}

Solution steps

4 x^{2} + 4 y^{2} + 24 x + 16 y + 27 = 0

Rewrite

4 x^{2} + 4 y^{2} + 24 x + 16 y + 27 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 3, - 2), r = \frac{5}{2}

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

x^{2} + y^{2} - 8 x - 2 y - 19 = 0

\frac{x ^{2}}{4} + \frac{y ^{2}}{4} = 1

16 + x^{2} + y^{2} - 8 x - 6 y = 0

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

What is 4x^2+4y^2+24x+16y+27=0 ?
The solution to 4x^2+4y^2+24x+16y+27=0 is Circle with (a,b)=(-3,-2),r= 5/2