What is x^2+24x+y^2+10y=-160 ?

The solution to x^2+24x+y^2+10y=-160 is Circle with (a,b)=(-12,-5),r=3

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x^2+24x+y^2+10y=-160

x^{2} + 24 x + y^{2} + 10 y = - 160

(a, b) = (- 12, - 5), r = 3

Solution steps

x^{2} + 24 x + y^{2} + 10 y = - 160

Rewrite

x^{2} + 24 x + y^{2} + 10 y = - 160

in the form of the standard circle equation

(x - (- 12))^{2} + (y - (- 5))^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (- 12, - 5), r = 3

x^{2} = 4 y

\frac{y ^{2}}{4} - \frac{x ^{2}}{5} = 1

6.25 x^{2} + 16 y^{2} - 12.5 x + 64 y - 29.75 = 0

4 y^{2} - 25 x^{2} = 100

- \frac{x ^{2}}{10} + \frac{9 x}{10} + \frac{11}{5}

What is x^2+24x+y^2+10y=-160 ?
The solution to x^2+24x+y^2+10y=-160 is Circle with (a,b)=(-12,-5),r=3