Implicit differentiation for circle

Question

Circle with equation $x^2+y^2-2x-2y+2 = 0$. When we do the implicit differentiation what we get is $\frac{dy}{dx} = \frac{1-x}{1-y}$, but what I noticed is that the radius of this cicle is zero, hence it is a point then what would this $\frac{dy}{dx}$ indicate?

Answer 1

Here are a couple of thoughts: first, suppose we try and compute $y' = \frac{dy}{dx}$ in the usual, elementary manner from the given implicit expression $x^2 + y^2 -2x - 2y + 2 = 0$; we obtain $2x - 2yy' -2 - 2y' = 0$ or, after a bit of re-arranging, $(x - 1) - y'(y - 1) = 0$. At this point, in order to isolate $y'$, a division by $y - 1$, if allowed, is required. However, it is easily seen that the given, original expression $x^2 + y^2 -2x - 2y + 2 = 0$ may be re-written as $(x - 1)^2 + (y - 1)^2 = 0$, which of course implies $x = y = 1$, whence $y - 1 = 0$ and hence division by $y - 1$ is in fact not permissible. So the expression $(x - 1) - y'(y - 1) = 0$ is as far as we can go in this direction; it is beginning to look like this is a case in which $y'$ can't really be defined. This observation is buttressed by the fact that any finite value we might assign to $y'$ will satisfy $(x - 1) - y'(y - 1) = 0$, so apparently $y'$ is not define-able under the given hypothesis.

To be somewhat blunt in answering the OP's question: $y' = \frac{dy}{dx}$ doesn't really indicate anything under these circumstances, since in cannot be assigned any sensible value.

One can also note that since $x = 1$ is forced upon us by the equation $x^2 + y^2 -2x - 2y + 2 = 0$, any usual definition of the derivative with respect to $x$ which relies on taking increments, viz $x \to x + {\Delta}{x}$, fails since $x$ is inherently non-incrementable here, i.e. we must have ${\Delta}{x} = 0$; thus any arguments based on limiting values of slopes of lines fall apart as well.

Finally, an amusing and hopefully instructional picture emerges if one looks at the more general equation $x^2 + y^2 -2x - 2y + 2 = R^2$ for real $R$; you can visualize it as a cone in $xyR$ space. Then as $R \to 0$ it appears that $y'$ (definable for $|x - 1| < |R|$) starts swinging about wildly (use your intuition to interpret this last, casual phrase); this behavior is an indicator that things break down at $R = 0$.

Hope these remarks help clarify the situation.

Answer 2

Well, if the circle is a point, then the only valid point where you can take the derivative is at $(1, 1)$, which would be that point. And, as you can notice, the derivative is $\frac{0}{0}$ at that point and doesn't exist.

What would a tangent line to a point be? Every line that passes through a point could be the tangent line to that point. The derivative here doesn't mean anything.