What does "the circle is tangent to the curve" mean? I've got a math exercise where it's said I have to prove that a circle is tangent to a curve (described by a parametric plot). Here's the graph :

So we can see that when $y=0$, the circle is really next to the curve. But it's not the case everywhere. My question is,  what do I have to prove in terms of limit to say that this circle is really tangent to the curve ? And is it tangent to it "everywhere"?
Thank you in advance. 
 A: 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. $f' = \infty$.
A: You can consider the tangent (straight) line tangent to the circle at a given point $(x_0,y_0)$. To say that the circle is tangent to the curve at the point $(x_0,y_0)$ is the same thing as the circle and the curve have a common point $(x_0,y_0)$ and at that point both the circle and the curve share the same tangent line. To find the tangent line to the circle (or to the curve) given by $y=f(x)$ at the point $(x_0,y_0)$ you have to check two conditions:


*

*The point belongs to the circle (or to the curve) so you must check that $y_0=f(x_0)$, 

*and to check that the equation of the line tangent to the circle (or to the curve) is given by $y=mx+b$ with $m=f'(x_0)$. If $f'(x_0)= \infty$ the slope is infinite so the tangent line is vertical. 
Edit: following the comment by J.M. if the curve is given parametrically by $x=x(t)$ and $y=y(t)$ then the equation of the curve is given by  (at the point $(x_0,y_0)=(x(t_0),y(t_0))$) : $y=y'(t_0)/x'(t_0) x+b$. Again if $x'(t_0)=0$ and $y'(t_0)\neq 0$ the slope is infinite and the line is vertical.
