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

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

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(x^2+4x)+(y^2-10y)=-4

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + y^{2} = (33)^{2}

(x - 4)^{2} + (y - 1)^{2} = 14

3 x^{2} + 3 y^{2} - 18 x + 6 y + 2 = 0

\frac{4 ( x + \frac{5}{2} ) ^{2}}{45} + \frac{4 ( y - 2 ) ^{2}}{45} = 1

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

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