Circles and lines geometry problem

Question

Here is my task: An arbitrary ray $OE$ crosses the circle at point $D$: $$x ^ 2 + (y - a / 2) ^ 2 = (a / 2) ^ 2$$ and a tangent to it(at point $E$), passing through point $C$, diametrically opposite to $O$.

Through points $D$ and $E$, straight lines are drawn parallel to the axes $Ox$ and $Oy$, respectively, up to the intersection at point M.

Equate the line formed by the $M$ points and draw it (Agnesi's curl).

My attempt

I think that this problem need using Polar coordinates. But I have no idea how to start. Maybe if we assume that $M$ has coordinates $(x,y)$ we can find this point(using polar coordinates like $x=r\sin(q)$ or something like that) but how to equate the line? And of course I have already drawn it. After an hour i have idea if we find point $M$ can we use $M$-coordinates to get parametric equation of a line? If yes, how to make first step? Ok, after a few time i found $x$ like a coordinate of a point $M$ (i use as a parameter angle between $OE$ and $OC$ (q) so $x=tgqa/cosq$) how can find $OB$?

Answer 1

Spoiler alert: this is not a hint, here is a solution using cartesian coordinates.

The calculation is based on similarity of triangles $OBD$ and $OCE$. First we consider the right side of the curve ($x \ge 0$) and then we note that the curve is symmetric. Our objective is to find $y_M$ as a function of $x_M$.

From the similarity of triangles $OBD$ and $OCE$ we have: $$\frac{OB}{OC} = \frac{BD}{CE} $$ $$\frac{y_M}{a} = \frac{x_D}{x_M} $$ $$\therefore y_M = a \frac{x_D}{x_M} \qquad(1)$$ Now, to find $x_D$ we note that $D$ is on the given circle and that $y_D = y_M$ : $${x_D}^2 + (y_M - \frac a2)^2 = (\frac a2)^2 $$ $$\therefore {x_D}^2 = y_M (a - y_M) \qquad(2)$$ By squaring equation (1) and using equation (2) we have: $$y_M = a^2 \frac{a-y_M}{{x_M}^2} $$ which can be rewritten as $$y_M = \frac{a^3}{a^2 + {x_M}^2} \qquad(3)$$ which is the equation of the loci of $M$ . Note that the curve of desired loci is symmetric, and equation (3) accounts for positive as well as negative values of $x$ .

Answer 2

In conventional polar coordinates $(\rho, \theta)$ we measure the angle $\theta$ counterclockwise from the $x$ axis. Of course you can instead use the angle $q$ measured clockwise from the $y$ axis, but then you must be careful not to simply copy any formulas that were developed for shapes in conventional polar coordinates. You have already correctly rederived at least one such formula, which is $x = r \cos\theta$ when converting conventional polar coordinates to Cartesian but is $x = r \sin q$ in your coordinates.

There are well-known formulas for a circle tangent to the origin in polar coordinates. You can derive a formula for your circle by substituting expressions with $r$ and $q$ for $x$ and $y$ in $$ x^2 + (y - a/2)^2 = (a / 2)^2, $$ or perhaps more easily by adding the segment $CD$ to the figure and examining the right triangle $\triangle ODC$ with right angle at $D.$

You should eventually be able to figure out both the $x$ and $y$ coordinates of $M$ using the angle $q$ as a parameter. That would be one way to describe the curve.

However, the fact that you have the formula for the circle in Cartesian coordinates suggests that someone was thinking of doing the whole problem in Cartesian coordinates. Using the $x$ coordinate of $M$ -- you might say the $x$ coordinate is $x_M$ -- you can write an equation of the line $OE$ in Cartesian coordinates. Then the coordinates of $D$ are what you get when you solve for $x$ and $y$ simultaneously in $ x^2 + (y - a/2)^2 = (a / 2)^2 $ and in the equation of the line $OE.$ What you're looking for is an equation relating the $x$ and $y$ coordinates of $M$.

So those are two ways of describing the curve. There may be other hints in the context of the question that would suggest which one will be more desirable.

Answer 3

Agnesi curve

Parametric equation

$\theta = angle(Ox, OD) \qquad OC = a$

triangle (OCE) : $CE = \dfrac{a}{\tan \theta} = x_{M}$

triangle (ODC) : $OD = OC\cos\Big( \dfrac{\pi}{2}-\theta \Big)= a\cos\Big( \dfrac{\pi}{2}-\theta \Big) = a\sin \theta$

triangle (OHD) : $HD = OD\cos\Big( \dfrac{\pi}{2}-\theta \Big)= OD\sin \theta = a\sin^{2}\theta = y_{M}$

equation : $x = \dfrac{a}{\tan \theta} \qquad y = a\sin^{2}\theta \qquad \theta \in [0, \pi]$