What is x^{(2)}+y^{(2)}+4x-10y=-16 ?

The solution to x^{(2)}+y^{(2)}+4x-10y=-16 is Circle with (a,b)=(-2,5),r=sqrt(13)

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x^{(2)}+y^{(2)}+4x-10y=-16

x^{(2)} + y^{(2)} + 4 x - 10 y = - 16

(a, b) = (- 2, 5), r = 13

Solution steps

x^{2} + y^{2} + 4 x - 10 y = - 16

Rewrite

x^{2} + y^{2} + 4 x - 10 y = - 16

in the form of the standard circle equation

(x - (- 2))^{2} + (y - 5)^{2} = (13)^{2}

Therefore the circle properties are:

(a, b) = (- 2, 5), r = 13

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

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

\frac{y ^{2}}{2} = - \frac{x ^{2}}{2} + 2

x^{2} + y^{2} + 10 x - 56 = 0

(x + 5)^{2} + (y + 2)^{2} = 21

What is x^{(2)}+y^{(2)}+4x-10y=-16 ?
The solution to x^{(2)}+y^{(2)}+4x-10y=-16 is Circle with (a,b)=(-2,5),r=sqrt(13)