# Find parameters

Find parameters of the circle in the plane - coordinates of center and radius:

$x2+(y-3)2=14$

Result

x0 =  0
y0 =  3
r =  3.742

#### Solution:

$(x-x_{ 0 })^2+(y-y_{ 0 })^2 = r^2 \ \\ (x-0)^2+(y-3)^2 = 14 \ \\ \ \\ x_{ 0 } = 0$
$y_{ 0 } = 3$
$r = \sqrt{ 14 } = \sqrt{ 14 } \doteq 3.7417 = 3.742$

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