A circle rests in the interior of the parabola with equation $y=x^2$ so that it is tangent to the parabola at two points. How much higher is the center of the circle than the points of tangency ?

I've seen the solution of the problem here on Aops site and it left me a bit skeptical.

I am not referring to the validity of the solution,which is correct, but rather to the fact that the problem has been solved using basic calculus.

Now, I came across this problem on Aops 2 volume book which doesn't teach any calculus,so it's clear that there's a simpler solution to the problem that relies only on basic geometry;however I've not been able to find it...

So I am asking if someone knows a solution which involves only basic geometry.


You don't need calculus for this one. Let the points of tangency be $(\pm a, a^2)$ and the center of the circle be $(0,c)$. Then the radius of the circle, $r$, is given by the Pythagorean theorem: $$r^2 = (c-a^2)^2 + a^2.$$ We can now plug this into the equation for the circle, $x^2 + (y-c)^2 = r^2$, to get: $$x^2 + y^2 - 2cy + c^2 - (c-a^2)^2 - a^2 = 0$$ but then we have $y=x^2$ so we get $$x^4 + (1-2c)x^2 + c^2 - a^2 - (c - a^2)^2 = 0.$$ We want the circle to only have two points of tangecy with the parabola, so we want the discriminant to $=0$, i.e. $$1 - 4c + 4c^2 - 8a^2c + 4a^4 + 4a^2 = 0$$ so that $$c = a^2 + \frac{1}{2}.$$ So now the answer to the question is simply $$a^2 + \frac{1}{2} - a^2 = \frac{1}{2} .$$

  • $\begingroup$ +1 but a typo in $r^2=a^4+(1-2c)a^2+c$, it should be $r^2=a^4+(1-2c)a^2+c^2$ $\endgroup$ – Ekaveera Kumar Sharma Dec 4 '15 at 13:33
  • $\begingroup$ Sorry about the typos, I think it's all cleaned up now. $\endgroup$ – TrivialCase Dec 4 '15 at 14:39

This problem has a name -- "The Quadratic Circle".

To solve it, we need knowledge slightly beyond simple geometry. They are :- (1) The theory of quadratic equation; and (2) Analytic geometry on circles.

enter image description here $y = x^2$ .............(1)

From given, $K$, the center of the unit circle must lie on the y-axis. Thus, $K= (0, k)$ for some $k$.

Then, the equation of that circle is $x^2 + (y – k)^2 = 1^2$ .............(2)

Putting (1) in (2), we have $y + (y – k)^2 = 1$

: :

$y^2 + (1 – 2k)y + [k^2 – 1] = 0$ ..........(3)

Let the points of intersection of the two curves be at $(x_1, y_1)$ and $(x_2, y_2)$.

By symmetry again, $y_1 = y_2$.

That means (3) has equal roots.

∴ $(1 – 2k)^2 – 4(1)[k^2 – 1] = 0$

: :

∴ $k = \dfrac {5}{4}$

  • $\begingroup$ Thanks for your answer @Mick $\endgroup$ – Mr. Y Dec 4 '15 at 19:38
  • $\begingroup$ @Mr.Y Hope that helps. $\endgroup$ – Mick Dec 5 '15 at 1:17

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