Hello Robin,
For a curve in the plane expressed as a function $y(x)$ in Cartesian coordinates, the radius of curvature is given by:
$$R(x)=\left|\frac{\left(1+y'^2 \right)^{\frac{3}{2}}}{y''} \right|$$
We are given the curve:
$$y(x)=\frac{1}{2}x^2$$
Hence:
$$y'=x$$
$$y''=1$$
And so the radius of curvature for this function is given by:
$$R(x)=(1+x^2)^{\frac{3}{2}}$$
Hence:
$$R(-1)=2\sqrt{2}$$
Now, the center of the osculating circle will lie aline the normal line at the given point. The slope $m$ of this normal line is the negative multiplicative inverse of the slope of the tangent line. Thus:
$$m=-\frac{1}{\left.y'(-1) \right|_{x=-1}}=-\frac{1}{-1}=1$$
Thus, using the point-slope formula, the normal line is given by:
$$y-\frac{1}{2}=x+1$$
Now, the distance from the tangent point of the osculating circle and its center $\left(x_C,y_C \right)$ is the radius of curvature we found above, and so we may write:
$$\left(x_C+1 \right)^2+\left(y_C-\frac{1}{2} \right)^2=\left(2\sqrt{2} \right)^2$$
Since the center of the circle lies on the normal line we found, we have:
$$\left(x_C+1 \right)^2+\left(x_C+1 \right)^2=8$$
$$\left(x_C+1 \right)^2=4$$
$$x_C=-1\pm2$$
Since the curve is concave up, we take the root:
$$x_C=1\implies y_C=\frac{5}{2}$$
Thus, the center of the osculating circle at the given point is:
$$\left(x_C,y_C \right)=\left(1,\frac{5}{2} \right)$$
