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 September 21st, 2013, 12:55 AM #1 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 circle inscribed inside parabolic sector Here is a fun little problem. Not crazy difficult, but more like a contest problem. "A parabolic sector is bounded by $y=-x^{2}+c$ and $y=0$. A circle of max area is inscribed in this sector. Find the radius of the circle in terms of c". Assume $c\geq 1$
 September 21st, 2013, 04:55 AM #2 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: circle inscribed inside parabolic sector The only obvious fact is that the circle must be centered on the y-axis. The first guess would be to set the radius to equal $\frac{c}{2}$. However, in this situation, the circle doesn't stay completely in the parabolic sector. In order to get the circle to be of maximal area and inscribed in the sector, then its boundary must be tangent to the x-axis and the parabola. Let the maximal circle have radius $r$, then its equation is $x^2 + (y - r)^2= r^2$. We will get an equation for the tangent vector to the circle at $(x, y(x))$ arbitrary. $x^2 + (y - r)^2= r^2 \ \Leftrightarrow \ y^2 - 2ry + x^2 = 0 \ \Rightarrow \ 2y(\frac{dy}{dx}) - 2r(\frac{dy}{dx}) + 2x = 0 .$ As such $\dfrac{dy}{dx}= \dfrac{x}{r - y}$ and a tangent vector to the circular boundary at $(x, y(x))$ is given by $\left( 1 , \dfrac{x}{r - y} \right)$. Likewise, a tangent vector to the parabola at $(x, y(x))$ is given by $(1, -2x)$. Where the parabola and circular boundary intersect, we must also have their tangents to coincide. Hence we have $-2x= \frac{x}{r - y} \ \Leftrightarrow \ y = \frac{1 + 2r}{2} \ \Rightarrow \ c - x^2 = \frac{1 + 2r}{2} \ \Rightarrow \ x^2 = c - \frac{1 + 2r}{2}$ At the 2 points of intersection; $-x^2 + c= r \pm \sqrt{r^2 - x^2} \ \Rightarrow \ x^4 + x^2(2r - 2c + 1) + c^2 - 2rc = 0 \ \Rightarrow \ (x^2 + r - c + \frac{1}{2})^2 = r^2 + r - c + \frac{1}{4}$. But from coincidence of the tangents at intersection points, we got $x^2= c - \frac{1 + 2r}{2}$ so that the left hand side directly above vanishes. In conclusion, $r^2 + r + \frac{1}{4} - c= 0 \ \Rightarrow \ (r + \frac{1}{2})^2 = c \ \Rightarrow \ r = \sqrt{c} - \frac{1}{2} .$ In conclusion, for $c \geq 1$, the radius of the inscribed circle must be $\sqrt{c} - \frac{1}{2} .$
 September 21st, 2013, 11:23 AM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: circle inscribed inside parabolic sector $x^2\,+\,y^2\,-\,2ry\,+\,r^2\,=\,r^2\,\Rightarrow\,x\,=\,\sqrt{2r y\,-\,y^2} \\ y\,=\,-x^2\,+\,c\,\Rightarrow\,x\,=\,\sqrt{c\,-\,y} \\ \sqrt{2ry\,-\,y^2}\,=\,\sqrt{c\,-\,y}\,\Rightarrow\,r\,=\,\frac{y^2\,-\,y\,+\,c}{2y}$ $y'\,=\,(x^2\,+\,y^2\,-\,2ry\,+\,r^2)'\,=\,\frac{x}{r\,-\,y} \\ y'\,=\,(-x^2\,+\,c)'\,=\,-2x \\ \frac{x}{r\,-\,y}\,=\,-2x\,\Rightarrow\,y\,=\,\frac{2r\,+\,1}{2}$ \begin{align*}r\,&=\,\frac{y^2\,-\,y\,+\,c}{2y} \\ r\,&=\,\frac{$$\frac{2r\,+\,1}{2}$$^2\,-\,\frac{2r\,+\,1}{2}\,+\,c}{2r\,+\,1} \\ r\,&=\,\frac{4r^2\,+\,4c\,-\,1}{8r\,+\,4} \\ 8r^2\,+\,4r\,&=\,4r^2\,+\,4c\,-\,1 \\ 4r^2\,+\,4r\,-\,4c\,+\,1\,&=\,0 \\ r\,&=\,\frac{-4\,+\,\sqrt{16\,+\,16(4c\,-\,1)}}{8} \\ r\,&=\,-\frac12\,+\,\frac{\sqrt{16(1\,+\,4c\,-\,1)}}{8} \\ r\,&=\,-\frac12\,+\,\frac{\sqrt{64c}}{8} r\,&=\,-\frac12\,+\,\sqrt{c}\end{align*}
 September 27th, 2013, 03:40 AM #4 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 Re: circle inscribed inside parabolic sector Very nice, G. Kind of a fun little problem. Not super tough, but cool nonetheless. I tackled similar to the way you done in your second approach. By considering the discriminant. Let the equation of the circle be $x^{2}+(y-a)^{2}=a^{2}$, where 'a' is the radius of said circle. $x^{2}+(-x^{2}+c-a)^{2}-a^{2}=0$ Because the circle and parabola must have equal roots, set the discriminant to 0. $(2a-2c+1)^{2}=4(1)(-2ac+c^{2})=0$ $4a^{2}+4a-4c+1=0$ $a=\sqrt{c}-1/2$
 September 3rd, 2015, 12:47 PM #5 Newbie   Joined: Sep 2015 From: USA Posts: 2 Thanks: 0 So what if c<1/4? Redoing this problem with a more general parabola y=-Ax^2 + C gives the radius of the circle with the maximum area to be R = sqrt(C/A) - 1/2A For any positive A and C it is possible to draw a circle under the parabolic sector. But the above radius goes negative when AC < 1/4. What am I missing?
 September 3rd, 2015, 01:45 PM #6 Newbie   Joined: Sep 2015 From: USA Posts: 2 Thanks: 0 OK I see some of it. The circle is tangent to the parabola at the focus same height as the focus of the parabola. If the focus occurs below the x axis this problem doesn't make sense. Hence the condition on 1/4A. The maximum area of a circle under a parabolic segment whose focus is beneath the axis is still beyond me.

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