Question

Find an equation for the circle of curvature of the curve r(t) = (2 ln(t))i - [t-1/t]j at the point (0, -2) , where t = 1

Answer 1

Is
$$
\large{
r(t)=\left (2\ln t,-t+\cfrac{1}{t}\right )\\
}
$$
?
Because
$$
r(1)=(0,0)
$$
and
$$
r(1)\ne(0,-2)
$$
Which is what you have above.

You'll need
$$
\large{
r'(t)=(x'(t),y'(t))\\
r''(t)=(x''(t),y''(t))
}
$$
To get
$$
\large{
\kappa(t)= \left|\frac{x'y''-y'x''}{({x'^2+y'^2})^{3/2}}\right|\\
}
$$
http://en.wikipedia.org/wiki/Curvature#Example