What is x^2+y^2-8x+10y=128 ?

The solution to x^2+y^2-8x+10y=128 is Circle with (a,b)=(4,-5),r=13

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x^2+y^2-8x+10y=128

x^{2} + y^{2} - 8 x + 10 y = 128

(a, b) = (4, - 5), r = 13

Solution steps

x^{2} + y^{2} - 8 x + 10 y = 128

Rewrite

x^{2} + y^{2} - 8 x + 10 y = 128

in the form of the standard circle equation

(x - 4)^{2} + (y - (- 5))^{2} = 1 3^{2}

Therefore the circle properties are:

(a, b) = (4, - 5), r = 13

2 x^{2} + 2 y^{2} - 3 x + 2 y - 10 = 0

(x - 3)^{2} + (y + 4)^{2} = 36

8 x + 32 y + y^{2} = - 263 - x^{2}

2 x^{2} + 2 y^{2} - 8 x + 28 y + 86 = 0

(x + 12)^{2} + (Y (x) - 4)^{2} - 30 = 0

What is x^2+y^2-8x+10y=128 ?
The solution to x^2+y^2-8x+10y=128 is Circle with (a,b)=(4,-5),r=13