Intersection of parabola and circle Is it possible to place circle and a parabola on the plane so that their intersection consist of exactly two points, one point being a point of tangency, and the other point a transversal intersection?
 A: First, thanks to Rahul Narain for finding a weak spot (aka error) in an earlier answer of mine.
Consider the standard parabola $y=x^2$, and the circle 
$$(x+4)^2 +(y-7/2)^2 =125/4.$$
Substituting $x^2$ for $y$, we find that the resulting quartic has a triple root at $x=1$, and a third root at $x=-3$. So the parabola and the circle are tangent at $(1,1)$. 
It is easy to check that we indeed have a triple root at $1$. The example was found in an  attempt to refute Rahul Narain's objection. Again, thanks.
A: Let the circle be the unit circle, and take the parabola to have the equation $y-k=a(x-h)^2$, with $a\neq 0$. Note that the slope of the line tangent to the parabola at $(x,y)$ is given by $m = 2 a (x-h)$.
Let $T(\cos\theta,\sin\theta)$ be the point of tangency. (We may assume throughout that $\sin\theta \neq 0$, since our (presumably non-degenerate) parabola has no vertical tangents. Note also that distinct points of intersection must correspond to distinct $x$ coordinates.) Then, as $T$ must satisfy the parabola equation, we have
$$\sin\theta - k = a ( \cos\theta - h )^2 \qquad (1)$$
Now, the line tangent to the circle at $T$ necessarily has slope $-\cot\theta$; equating this with the slope relative to the parabola gives
$$\cos\theta = -2 a \sin\theta (\cos\theta - h ) \qquad (2)$$
We can solve $(2)$ for $h$, and then $(1)$ for $k$:
$$
h = \frac{\cos\theta \left( 1 + 2 a \sin\theta \right)}{2a\sin\theta}
\qquad
k = \frac{ 4 a \sin^3\theta - \cos^2\theta }{4a\sin\theta^2}$$
whence the parabola equation becomes
$$y \sin\theta = a x^2 \sin\theta - x \cos\theta \left( 1 + 2 a \sin\theta \right) + 1 + a \cos^2\theta \sin\theta $$
Using that equation to eliminate $y$ in $x^2+y^2=1$ gives this polynomial equation in $x$:
$$\left( x - \cos\theta \right)^2 \left(
 \left( a \left( x - \cos\theta \right)\sin\theta - \cos\theta\right)^2
 + \sin^2\theta \left( 1 + 2 a \sin\theta \right)\right) = 0 \qquad (3) $$
For the second factor of $(3)$ to have any roots requires the quantity $b := 1+2a\sin\theta$ to be non-positive. If $b$ vanishes, we get the double root $x=-\cos\theta$ (the mirror image of $x=\cos\theta$) corresponding to a second point of circle-parabola tangency. For us to have a point of non-tangent intersection, we require $b$ to be strictly negative, and $x=\cos\theta$ to be a root of that second factor. The latter condition implies, by substitution:
$$2a\sin^3\theta=-1$$
giving $b=-\cot^2\theta$, which is strictly negative (and defined) for $\theta \neq n \pi/2$. Moreover, equation $(3)$ reduces nicely to
$$0 = (x-\cos\theta)^3\left(x-4\cos^3\theta+3\cos\theta\right) = (x-\cos\theta)^3(x-\cos 3\theta)$$
providing the root $x=\cos3\theta$. Double-checking with the parabola equation, we see that the corresponding $y$ value should be $-\sin3\theta$; thus, we can write the second point of intersection as $S(\cos(-3\theta), \sin(-3\theta))$.
The final parabola equation itself is
$$ 2 y \sin^3\theta = - \left( x - \cos^3\theta \right)^2 + \sin^4\theta\left( 3 - \sin^2\theta \right)$$
Here's an animation:

(Animation, and interstitial symbol-crunching, courtesy of Mathematica.)
A: Just on a different flavour. Let the parabola be the standard $Y=X^2$ (which you can always assume by properly chosing coordinates).
Fix a point $P$ on this parabola. 
By using complex projective coordinates you can generate the linear system of conics which are:


*

*tangent to the parabola in $P$;

*passing through the cyclic (improper) points $(1,\imath,0)$, $(1,,-\imath,0)$;


Such linear system is generated by two degenerate conics (pairs of lines).
Each conic (apart from degenerate ones) in this linear system is a circle. The condition of passing through an arbitary point in the plane, in particular a point lying on the parabola, picks one.
This implies that for any parabola and any pair of distinct points on it there is a solution to the problem. Algebraic computations are easy
