What is x^2+y^2+4x-6y+12=0 ?

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

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

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

(a, b) = (- 2, 3), r = 1

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 2, 3), r = 1

x^{2} = 16 y

f (x) = - x^{2} + 6 x + 7

- \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

\frac{x ^{2}}{16} - \frac{y ^{2}}{4} = 1

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

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