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

The solution to x^2-4x+y^2+6y-12=0 is Circle with (a,b)=(2,-3),r=5

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

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

(a, b) = (2, - 3), r = 5

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (2, - 3), r = 5

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

x^{2} = - 4 y

y^{2} = 4 - x^{2}

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

\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1

What is x^2-4x+y^2+6y-12=0 ?
The solution to x^2-4x+y^2+6y-12=0 is Circle with (a,b)=(2,-3),r=5