What is x^2+y^2-8x+4y=0 ?

The solution to x^2+y^2-8x+4y=0 is Circle with (a,b)=(4,-2),r=2sqrt(5)

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x^2+y^2-8x+4y=0

x^{2} + y^{2} - 8 x + 4 y = 0

(a, b) = (4, - 2), r = 25

Solution steps

x^{2} + y^{2} - 8 x + 4 y = 0

Rewrite

x^{2} + y^{2} - 8 x + 4 y = 0

in the form of the standard circle equation

(x - 4)^{2} + (y - (- 2))^{2} = (25)^{2}

Therefore the circle properties are:

(a, b) = (4, - 2), r = 25

(x + 4)^{2} + (y - 4 \frac{1}{2})^{2} = 42 \frac{1}{4}

x^{2} + y^{2} + 8 x - 2 y + 13 = 0

x^{2} + y^{2} - 6 y + 8 = 0

x^{2} + y^{2} - 10 x - 14 y + 58 = 0

4 x^{2} + 4 y^{2} - 48 x + 20 y = - 141

What is x^2+y^2-8x+4y=0 ?
The solution to x^2+y^2-8x+4y=0 is Circle with (a,b)=(4,-2),r=2sqrt(5)