# 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"?

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It means at the point where both curves meet, they have the same slope... – Listing Oct 26 '11 at 13:03
I would say: compute tangent lines to both of them at that point, and see if it is the same line. No, it is just tangent at one (or, in this case, maybe two) point. – GEdgar Oct 26 '11 at 13:24

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

1. there is $t_1,t_2$ s.t. $z = (x_1(t_1),y_1(t_1)) = (x_2(t_2),y_2(t_2))$;

2. functions $x_i,y_i$ are differentiable at $t_i$ and their derivatives coincide: $$\dot{x}_1(t_1) = \dot{x}_1(t_2) \text{ and }\dot{y}_1(t_1) = \dot{y}_2(t_2).$$

3. 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. $f' = \infty$.

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Thank you too. I didn't need so many information because all is not of my level but that's good :) – Skydreamer Oct 26 '11 at 13:22
@Skydreamer You're welcome. Btw, be aware that re-accepting answers is not the policy people will like. To avoid it you may want not to accept the answer 1 hour after you asked your question. Or do not change you decision if you've done it. Just an advice. – Ilya Oct 26 '11 at 13:27
I'm sorry, I didn't know it :) – Skydreamer Oct 26 '11 at 13:38

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:

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

2. 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.

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In the example given by the asker, the tangent line is vertical, so $m$ is undefined. – Quinn Culver Oct 26 '11 at 13:11
Also, OP said that the curve was given parametrically, so a better form for the equation of the tangent line goes something like $f^\prime(t)(y-g(t))=g^\prime(t)(x-f(t))$... – J. M. Oct 26 '11 at 13:14
Thank you for your nice answer :) – Skydreamer Oct 26 '11 at 13:16