What is 0=1-13(x^2+y^2) ?

The solution to 0=1-13(x^2+y^2) is Circle with (a,b)=(0,0),r=sqrt(1/13)

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0=1-13(x^2+y^2)

0 = 1 - 13 (x^{2} + y^{2})

(a, b) = (0, 0), r = \frac{1}{13}

Solution steps

0 = 1 - 13 (x^{2} + y^{2})

Rewrite

0 = 1 - 13 (x^{2} + y^{2})

in the form of the standard circle equation

(x - 0)^{2} + (y - 0)^{2} = (\frac{1}{13})^{2}

Therefore the circle properties are:

(a, b) = (0, 0), r = \frac{1}{13}

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

4 x^{2} + 4 y^{2} - 24 x - 8 y + 15 = 0

x^{2} + y^{2} + 2 x - 10 y - 8.6 = 0

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

(x + 4)^{2} + (y + 3)^{2} = 11

What is 0=1-13(x^2+y^2) ?
The solution to 0=1-13(x^2+y^2) is Circle with (a,b)=(0,0),r=sqrt(1/13)