What is the radius (x+1/3)^2+(y-5)^2=sqrt(7) ?

The radius (x+1/3)^2+(y-5)^2=sqrt(7) is r=sqrt(\sqrt{7)}

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radius (x+1/3)^2+(y-5)^2=sqrt(7)

radius

(x + \frac{1}{3})^{2} + (y - 5)^{2} = 7

r = 7

Solution steps

(x + \frac{1}{3})^{2} + (y - 5)^{2} = 7

Rewrite

(x + \frac{1}{3})^{2} + (y - 5)^{2} = 7

in the form of the standard circle equation

(x - (- \frac{1}{3}))^{2} + (y - 5)^{2} = (7)^{2}

Therefore the circle properties are:

(a, b) = (- \frac{1}{3}, 5), r = 7

And the radius is:

r = 7

x^{2} + y^{2} + 8 x - 12 y + 27 = 0

x^{2} + (y - 3)^{2} = 16

(x + 2)^{2} + (y - 5)^{2} = 16

(x - 2)^{2} + (y + 4)^{2} = 9

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

What is the radius (x+1/3)^2+(y-5)^2=sqrt(7) ?
The radius (x+1/3)^2+(y-5)^2=sqrt(7) is r=sqrt(\sqrt{7)}