A bob of mass $m$ and negligible radius is tied to a massless string of length $L$ and rotated uniformly along a vertical circle of radius $L$ under the influence of gravity. The free end of the taut string has to be moved around in the plane of the circle so that the speed of the bob never changes. Find the locus of the free end of the string.
My work: Assume the vertical circle to be $x^{2}+y^{2}=L^{2}$, and let the bob's position be modeled by $\left(L\cos\theta,L\sin\theta\right)$, where $\theta=\omega t$. Let the free end of the string be positioned such that the line joining the free end and the bob; and the bob and the origin make an angle $\varphi$. Obviously, the free end of the string, the bob and the origin form an isosceles triangle with equal sides $L$.
Now to write the equations. To prevent any force acting in the tangential direction, $T\sin\varphi=mg\cos\theta$ and to avoid force in the radial direction, $T\cos\varphi+mg\sin\theta=\frac{mv^{2}}{L}$ where $v=L\omega$.
On solving, we get $$\varphi=\cot^{-1}\left(\frac{v^{2}-Lg\sin\theta}{Lg\cos\theta}\right)$$ Now the position of the free end of the string can be given by $\left(R\cos\alpha,R\sin\alpha\right)$, where $R=2L\sin\left(\frac{\varphi}{2}\right)$ (from cosine rule) and $\alpha=\frac{\pi}{2}-\frac{\varphi}{2}+\theta$.
How would you get the locus of this point? I'm absolutely stumped since it looks like a lot of parameters to handle. Thanks in advance for any help!
Attaching an image, if it helps.
You have $$\varphi=\cot^{-1}\left(\frac{v^{2}-Lg\sin\theta}{Lg\cos\theta}\right)\tag1$$ $$x=R\cos\alpha\tag2$$ $$y=R\sin\alpha\tag3$$ $$R=2L\sin\frac{\varphi}{2}\tag4$$ $$\alpha=\frac{\pi}{2}-\frac{\varphi}{2}+\theta\tag5$$
From $(2)(4)(5)$, we have $$\begin{align}x&=2L\sin\frac{\varphi}{2}\cos\bigg(\frac{\pi}{2}-\frac{\varphi}{2}+\theta\bigg) \\\\&=-2L\sin\frac{\varphi}{2}\sin\bigg(\theta-\frac{\varphi}{2}\bigg) \\\\&=-2L\sin\frac{\varphi}{2}\bigg(\sin\theta\cos\frac{\varphi}{2}-\cos\theta\sin\frac{\varphi}{2}\bigg) \\\\&=-L\sin\theta\sin\varphi+L\cos\theta(1-\cos\varphi)\end{align}$$ So, we get $$L\sin\theta\sin\varphi+L\cos\theta\cos\varphi=L\cos\theta-x\tag6$$
From $(3)(4)(5)$, we have $$\begin{align}y&=2L\sin\frac{\varphi}{2}\sin\bigg(\frac{\pi}{2}-\frac{\varphi}{2}+\theta\bigg) \\\\&=2L\sin\frac{\varphi}{2}\cos\bigg(\theta-\frac{\varphi}{2}\bigg) \\\\&=2L\sin\frac{\varphi}{2}\bigg(\cos\theta\cos\frac{\varphi}{2}+\sin\theta\sin\frac{\varphi}{2}\bigg) \\\\&=L\cos\theta\sin\varphi+L\sin\theta(1-\cos\varphi)\end{align}$$
So, we get $$L\cos\theta\sin\varphi-L\sin\theta\cos\varphi=y-L\sin\theta\tag7$$
Solving $(6)(7)$ for $\cos\varphi$ and $\sin\varphi$, we have $$\cos\varphi=\frac{L-x\cos\theta-y\sin\theta}{L}\tag8$$ $$\sin\varphi=\frac{-x\sin\theta+y\cos\theta}{L}\tag9$$
From $(1)$, we have $$\frac{\cos\varphi}{\sin\varphi}=\frac{v^{2}-Lg\sin\theta}{Lg\cos\theta}$$
So, from $(8)(9)$, we get $$\frac{L-x\cos\theta-y\sin\theta}{-x\sin\theta+y\cos\theta}=\frac{v^{2}-Lg\sin\theta}{Lg\cos\theta}$$ which can be written as $$(L^2g-v^2y)\cos\theta+v^2x\sin\theta=xLg\tag{10}$$
From $(2)(3)(4)$, we have $$\begin{align}x^2+y^2&=R^2 \\\\&=\bigg(2L\sin\frac{\varphi}{2}\bigg)^2 \\\\&=4L^2\sin^2\frac{\varphi}{2} \\\\&=2L^2(1-\cos\varphi)\end{align}$$
So, from $(8)$, we get $$x^2+y^2=2L^2-2L(L-x\cos\theta-y\sin\theta)$$ which can be written as $$2Lx\cos\theta+2Ly\sin\theta=x^2+y^2\tag{11}$$
Solving $(10)(11)$ for $\cos\theta$ and $\sin\theta$, we get $$\cos\theta=\frac{2L^2xyg-v^2x(x^2+y^2)}{2Ly(L^2g-v^2y)-2Lv^2x^2}$$ $$\sin\theta=\frac{(L^2g-v^2y)(x^2+y^2)-2x^2L^2g}{2Ly(L^2g-v^2y)-2Lv^2x^2}$$
Eliminating $\theta$, we finally get $$\bigg(\frac{2L^2xyg-v^2x(x^2+y^2)}{2Ly(L^2g-v^2y)-2Lv^2x^2}\bigg)^2+\bigg(\frac{(L^2g-v^2y)(x^2+y^2)-2x^2L^2g}{2Ly(L^2g-v^2y)-2Lv^2x^2}\bigg)^2=1$$ which can be written as
$$ \color{red}{v^4(x^2+y^2)^3- 2gL^2v^2(x^2+y^2)^2y + L^2(g^2 L^2 - 4 v^4)(x^2+y^2)^2 + 8 g L^4 v^2( x^2 + y^2) y -4 g^2 L^6 y^2 =0}$$