What is (x^2-4x)+(y^2+8y)=16 ?

The solution to (x^2-4x)+(y^2+8y)=16 is Circle with (a,b)=(2,-4),r=6

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(x^2-4x)+(y^2+8y)=16

(x^{2} - 4 x) + (y^{2} + 8 y) = 16

(a, b) = (2, - 4), r = 6

Solution steps

x^{2} - 4 x + (y^{2} + 8 y) = 16

Rewrite

x^{2} - 4 x + (y^{2} + 8 y) = 16

in the form of the standard circle equation

(x - 2)^{2} + (y - (- 4))^{2} = 6^{2}

Therefore the circle properties are:

(a, b) = (2, - 4), r = 6

x^{2} + y^{2} - 6 x + 10 y + 27 = 0

x^{2} + y^{2} - 12 x - 12 y + 36 = 0

(x + 3)^{2} + (y - 2)^{2} = 36

4 x^{2} + 4 y^{2} - 64 = 0

x^{2} + 2 x \cdot 3 + 3^{2} + y^{2} - 2 y \cdot \frac{2}{3} + (\frac{2}{3})^{2} = 4^{2}

What is (x^2-4x)+(y^2+8y)=16 ?
The solution to (x^2-4x)+(y^2+8y)=16 is Circle with (a,b)=(2,-4),r=6