What is x^2+y^2-4y-5=0 ?

The solution to x^2+y^2-4y-5=0 is Circle with (a,b)=(0,2),r=3

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x^2+y^2-4y-5=0

x^{2} + y^{2} - 4 y - 5 = 0

(a, b) = (0, 2), r = 3

Solution steps

x^{2} + y^{2} - 4 y - 5 = 0

Rewrite

x^{2} + y^{2} - 4 y - 5 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - 2)^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (0, 2), r = 3

- y^{(2)} + 14 y = x^{(2)} - 10 x + 25 + 49 - 12.24

\frac{x ^{2}}{4} + \frac{z ^{2}}{4} = 1

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

x^{2} + 4 x + y^{2} + 8 y - 5 = 0

x^{2} - 2 x + y^{2} - 15 = 0

What is x^2+y^2-4y-5=0 ?
The solution to x^2+y^2-4y-5=0 is Circle with (a,b)=(0,2),r=3