Largest possible radius of a circle lying in a parabola

Question

Circle of radius $r$ touches the parabola $y^2+12x=0$ at its vertex. Centre of circle lies left of the vertex and circle lies entirely within the parabola. What is the largest possible value of $r$?

So my book has given the solution as follows:

The equation of the circle can be taken as: $(x+r)^2+y^2=r^2$
and when we solve the equation of the circle and the parabola, we get $x=0$ or $x=12-2r$.

Then, $12-2r≥0$ and finally, the largest possible value of $r$ is $6$.

This is where I got stuck as I'm not able to understand why that condition must be true. I get that the circle must lie within the parabola...

Can someone please explain this condition to me?

Answer 1

For the circle to be entirely within the parabola means the parabola and the circle intersect at exactly one point.

However solving $(x+r)^2 + y^2 = r^2$ and $y^2 + 12x = 0$ we actually will find three (potential) points of intersection.

$y^2=-12x$ so $(x+r)^2 +y^2 =(x+r)^2 -12x = r^2$ and $x(x+2r -12) =0$ so

$x = 0$ and $y=0$, or $x = (12-2r)$ and $y = \pm \sqrt{12(2r-12)}$.

$x=0, y=0$ is okay, but we can't have $x = (12-2r)$ and $y= \pm \sqrt{12(2r-12)}$ be two more points of intersection.

The only we to avoid those two potential points existing is if either: i) $(12-2r, \pm \sqrt{12(2r-12)}) = (0,0)$; or in other words if $r=6$; or ii) if $\sqrt{12(2r-12)}$ doesn't exist; or in other words if $12(2r-12) < 0$ or in other words if $r < 6$.

So if $r \le 6$ then only one point of intersection exists.

If $r > 6$ then three points of intersection exist.

As we figure that three points of intersection was impossible, we concluder $r < 6$>

......

Hmm, I suppose we have to justify my very first sentence "For the circle to be entirely within the parabola means the parabola and the circle intersect at exactly one point".

Note: if $r > 6$ and if $2r -12 < x < 0$ then the points of the circle $y=\pm\sqrt{r^2 - (x+r)^2}$ will be so that $y^2 < -12x$ so those points of the circle are outside the parabola.

========

Maybe a clear explanation of what algebraically what a circle being within a parabola means.

A point $(w,v)$ is "inside" the parabola $y^2 + 12x =0$ or $y = \pm \sqrt {-12x}$ if the point $(w,v)$ is left of the vertex of $y^2 + 12x = 0$; that is to say that $w < 0$, and if the point $(w,v)$ is withing the bounds of the parabola at $x = w$; that is to say that if $y^2 + 12w =0$ so $y =\pm \sqrt {-12w}$ then the value of $v$ is so that $-\sqrt{-12w} \le v \le \sqrt{-12w}$.

Now the equation of the circle is $(x+r)^2 + y^2 = r^2$ so if we have a point $(w,v)$ on the circle that point is satisfies $(w+r)^2 + v^2 = r^2$ and if the circle is inside the parabola we must have $-\sqrt{-12w} \le v \le \sqrt{-12w}$.

Solving: $w^2 + 2wr +r^2 + v^2 = r^2$ so $v =\pm\sqrt{-w(w+2r)}$ so we need $-\sqrt{-12w} \le -\sqrt{-w(w+2r)} \le \sqrt{-w(w+2r)}\le \sqrt{-12w}$.

For all points of the circle we have $w$ between $-2r, 0$ so $w \le 0$. If $w=0$ we have $v=0$ and the inequality holds.

If $w < 0$ then $\sqrt{-w} > 0$ and we can divide the inequality be $\sqrt{-w}$ to get

$-\sqrt{12}\le -\sqrt{w+2r} \le \sqrt{w+2r} \le \sqrt{12}$ or $0 \le w+2r \le 12$. for all $w < 0$. That means that $2r < 12 + |w|$ for all $|w|>0$ or that $r < 6 + \epsilon$ for all $\epsilon > 0$.

Sigh... okay, this throws everyone the first time the see it but that means $r \le 6$. (If $r > 6$ then if we set $\epsilon = r-6$ we have $\epsilon > 0$ and we get the contradiction: $r < 6 + \epsilon = 6 +(r-6) = r$.)

.....

Hmmm.... this proof is nowhere near as clear or easy to follow as my first argument that if a circle is within a parabola that means it intersects at exactly one point.

But.... although it is geometrically obvious that that statement about intersection is true, I'm not sure we can justify it without an argument.

Answer 2

First of all, you have a typo, the circle is $ (x + r)^2 + y^2 = r^2 $.

Clearly since $ x \leq 0 $ on the parabola, any intersections of the circle with the parabola will have $ x \leq 0 $. $ x = 0 $ will always correspond to the vertex. The question is asking you to set the radius $ r $ as big as possible without having any other intersection points. Through the calculations you/they have determined that any other intersection points can only occur when $ x = 12 - 2r $ or $ x = 0 $. But since $ y^2 $ has to be non-negative ($ y $ is real after all) and we have $ y^2 = -12x $, $ x = 12 - 2r $ corresponds to an intersection point if and only if $ x \leq 0 $. So we get three intersection points if and only if $ 12 - 2r < 0 $, in which case the three intersection points are $ (0,0) $, $ \left(12 - 2r, \sqrt{12(2r-12)} \right) $, and $ \left(12 - 2r, -\sqrt{12(2r-12)} \right)$. Put differently, the circle is in the parabola if and only if there is (at most) one intersection point, which in turn is true if and only if $ r < 6 $.

Clearly if you have three intersection points, the circle is not "inside" the parabola.

Answer 3

$(x+r)^2+y^2=r^2$ (check the equation)

$y^2 + 12x = 0$

Equating the two,

$x^2 -12x + 2xr = 0$

$x^2 + (2r-12)x = 0$

So two solutions in $x$, $x = 0, (12-2r)$

For any value of $r \gt 6$, $x = 12 - 2r$ is negative and so you have another value of $x$ other than $0$ that satisfies the parabola equation $y^2 = -12x$ which means the circle will intersect the parabola for $r \gt 6$.

Note that for $r \lt 6$, $x = 12-2r$ is positive which cannot be on the parabola $y^2 = -12x$.

Answer 4

It is helps to learn and remember standard forms of the parabola along with the osculating circle radii at their vertices. $$ y^2= \pm 2Rx , x^2=\pm 2R y$$ Differentiating $ 2 R y = x^2$ twice show that the minimum touching circle radius is R.. in this case $6$.

Answer 5

Under given constraints, the maximal value of the radius $r_{\max}=\tfrac1\kappa$, where $\kappa$ is the curvature of the parabola at its vertex.

For the given parabola $y^2=-12x$ we can use a convenient parametric representation

\begin{align} y(t)&=t ,\quad x(t)=-\tfrac1{12}\,t^2 ,\\ y'(t)&=1 ,\quad x'(t)=-\tfrac1{6}\,t ,\\ y''(t)&=0 ,\quad x''(t)=-\tfrac1{6} , \end{align}

so we can use a known expression for the curvature of parametric curve

\begin{align} \kappa(t)&= \frac{x'y''-y'x''}{\sqrt{(x'^2+y'^2)^3}} \\ &=\frac{-\tfrac1{6}\,t\cdot0-1\cdot(-\tfrac16)} {\sqrt{\Big(\tfrac1{36}\,t^2+1\Big)^3}} =\frac{36}{\sqrt{(t^2+36)^3}} . \end{align}

The vertex is located at the point $(x(t),y(t))|_{t=0}$, so the answer is

\begin{align} r_{\max}&=\frac1{\kappa(0)} =\frac{\sqrt{(0^2+36)^3}}{36} =6 . \end{align}