What is x^2+y^2-7y+8=0 ?

The solution to x^2+y^2-7y+8=0 is Circle with (a,b)=(0, 7/2),r=(sqrt(17))/2

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x^2+y^2-7y+8=0

x^{2} + y^{2} - 7 y + 8 = 0

(a, b) = (0, \frac{7}{2}), r = \frac{17}{2}

Solution steps

x^{2} + y^{2} - 7 y + 8 = 0

Rewrite

x^{2} + y^{2} - 7 y + 8 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - \frac{7}{2})^{2} = (\frac{17}{2})^{2}

Therefore the circle properties are:

(a, b) = (0, \frac{7}{2}), r = \frac{17}{2}

x^{2} + 6 x + y^{2} - 10 y = 135

z^{2} = 4 - y^{2}

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

(x + 1)^{2} + (y - 5)^{2} = 20

x^{2} + y^{2} + 6 x - 8 y + 13 = 0

What is x^2+y^2-7y+8=0 ?
The solution to x^2+y^2-7y+8=0 is Circle with (a,b)=(0, 7/2),r=(sqrt(17))/2