I guess in analysis the usual application of 'tangency' is local, i.e. two curves $\gamma_1,\gamma_2$ with at least one common point $z\in \gamma_1\cap\gamma_2$ are said to be tangent at $z$ if their tangent lines coincides.
E.g. when $\gamma_i$ are given on the plain and parametrized by $ \gamma_i =\left\{(x_i(t),y_i(t)):t\in [a,b]\right\} $ for $i=1,2$ they are tangent at $z\in \mathbb R^2$ if
there is $t_1,t_2$ s.t. $z = (x_1(t_1),y_1(t_1)) = (x_2(t_2),y_2(t_2))$;
functions $x_i,y_i$ are differentiable at $t_i$ and their derivatives coincide: $ \dot{x}_1(t_1) = \dot{x}_1(t_2) \text{ and }\dot{y}_1(t_1) = \dot{y}_2(t_2). $
Derivatives are not degenerate: $\dot{x}_i^2(t_i)+\dot{y}_i^2(t_i)>0$ for $i=1,2.$ This assumption is made to avoid undesired 'corner' behavior.
Curves on your picture seem to be tangent, but to be sure you should take e.g. polar parameters for the circle: $ (x_1(t),y_1(t)) = (\cos{t},\sin{t}) $ with $t\in [0,2\pi]$ as well as polar parameters for the second curve. That curve seem to be given in polar coordinates $(r,\phi)$ - then you take $ (x_2(t),y_2(t)) = (r(t)\cos{t},r(t)\sin{t}) $ as parameters.
Be aware that answer by Ali may not work in your case since the slope of your curves at the points of intersection is vertical, i.e. f' = \infty.