On the plane is drawn a triangle $ABC$ and a circle $\omega$ passing through the vertex $C$, the midpoints of the sides $AC$ and $BC$ and the point of intersection of the medians of the triangle $ABC$. The point $K$ lies on the circle $\omega$ such that $\angle AKB=90^o$. Using only with a ruler, draw a tangent to the circle $\omega$ at point $K$.