What is x^2+y^2-13y+41=0 ?

The solution to x^2+y^2-13y+41=0 is Circle with (a,b)=(0, 13/2),r=(sqrt(5))/2

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x^2+y^2-13y+41=0

x^{2} + y^{2} - 13 y + 41 = 0

(a, b) = (0, \frac{13}{2}), r = \frac{5}{2}

Solution steps

x^{2} + y^{2} - 13 y + 41 = 0

Rewrite

x^{2} + y^{2} - 13 y + 41 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - \frac{13}{2})^{2} = (\frac{5}{2})^{2}

Therefore the circle properties are:

(a, b) = (0, \frac{13}{2}), r = \frac{5}{2}

x^{2} + y^{2} + 2 x - 3 y - 7 = 0

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

x^{2} + y^{2} - 5 x = 36 - 6 y

(x - 4)^{2} + (y - 6)^{2} = 12

x^{2} + y^{2} - 10 x + 12 y - 39 = 0

What is x^2+y^2-13y+41=0 ?
The solution to x^2+y^2-13y+41=0 is Circle with (a,b)=(0, 13/2),r=(sqrt(5))/2