Locus of the center of a semi-circle tangent to a fixed semicircle, both contained in a circle Following with a curious result of
the Cut-the-Knot entry "A Property of Semicircles:
What is this about? A Mathematical Droodle",

Let's suppose $AB$ is fixed diameter of one semicircle, but $CD$ (as a diameter) is a dynamic semicircle which always is tangent with semicircle ($AB$ as diameter). What's the track function will it be for point $N$ (middle point of $CD$)?
 A: 
The shortest possible answer to your question is “the Limaçon of Etienne Pascal”. This also known as the conchoid of a circle and its equation in the polar coordinate system is given by
$$r=a+b\cos\left(\theta\right). \tag{1}$$
However, the midpoint $N$ describes only the part of this quartic curve in the range $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$. The curve called cardioid appears as a special case of the Pascal limaçon when $a = b$.
Let $r_0$ be the radius of the semicircle $AQPB$, which is inscribed in the circle $\Gamma$ having radius $R$ and the point $O$ as its center (see $\mathrm{Fig.\space 1}$). Please note that the origin of the Cartesian coordinate system is located at $M$, the center of this semicircle. Let $r_1$ and $G$ be the radius and the center of the semicircle $EQF$, which is also inscribed the circle $\mathit{\Gamma}\space$ such that the two semicircles externally tangential to each other at $Q$.
The auxiliary line $MG$ is drawn to join the centers of the two semicircles and it is obvious that  it goes through $Q$ as it is their common point. Furthermore, the auxiliary lines $OG$ and $OF$ were also drawn. Since $G$ is the midpoint of $FE$, which is a chord of $\Gamma$, $OG$ is its perpendicular bisector. Let $MG$ and $\measuredangle OMG$ be $r$ and $\theta$ respectively, where $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$.
We start our search for the locus of $G$ by expressing $r_1$ in terms of $r$ and $r_0$.
$$MG = r_0 + r_1 = r \quad\rightarrow\quad r_1 = r\enspace –\space r_0 \tag{2}$$
Since $OE$ is a radius of $\Gamma$, $OE = R$. Likewise, since $GE$ is a radius of the semicircle $EQF$, $GE = r_1$. The triangle $OGE$ is a right angle triangle. Therefore,
$$OG^2 = R^2 - r_1^2. \tag{3}$$
The triangle $BMO$ is also a right angle triangle, hence, we shall write,
$$OM^2= R^2 – r_0^2. \tag{4}$$
We apply cosine law to $\triangle OMG$ to obtain,
$$OG^2=OM^2+MG^2-2\times OM\times MG\times \cos\left(\theta\right).\tag{5}$$
From (3), (4), and (5), it follows,
$$R^2 – r_1^2 = R^2 – r_0^2 + r^2 – 2r\sqrt{ R^2 – r_0^2}\cos\left(\theta\right).\tag{6}$$
Once we simplify the equation, which we obtained by eliminating $r_1$ from (6) using (2), we have,
$$r = r_0 + \sqrt{ R^2 – r_0^2}\cos\left(\theta\right). \tag{7}$$
This is the equation in polar form of the locus of both $G$ and $N$. This equation resembles (1). Therefore, we can conclude that the sought locus is the part of a Pascal limaçon that traverses from  $A\space\left(\theta =-\frac{\pi}{2}\right)\space$ to $\space B\space\left(\theta=\frac{\pi}{2}\right)$.
The $x-$ and $y-$coordinates of the midpoint $G$ can be expressed as,
$$x_G = r\cos\left(\theta\right) = r_0\cos\left(\theta\right) + \sqrt{ R^2 – r_0^2}\cos^2\left(\theta\right) \quad\quad\mathrm{and}\quad$$
$$y_G = \pm r\sin\left(\theta\right) = \pm \Bigl[ r_0\sin\left(\theta\right) + \sqrt{ R^2 – r_0^2}\cos\left(\theta\right) \sin\left(\theta\right)\Bigr].$$
The equation of the locus in the Cartesian coordinate system is given by
$$\left(x^2+y^2-x \sqrt{ R^2 – r_0^2}\right)^2=r_0^2 \left(x^2+y^2\right). \tag{8}$$
It is evident from (8) that the locus of $N$ and $G$ is a quartic (i.e. fourth order) curve.
We would like to present some additional information for OP’s perusal, which would answer the question “How does one construct the semicircles, such as $FEQ$ or $CDP$ shown in $\mathrm{Fig.\space 1}$, when the circle $\it\Gamma$ and the inscribed semicircle with the fixed radius are given?”
In the first scenario illustrated in $\mathrm{Fig.\space 2}$, we are given $\measuredangle \theta$ as well. Draw a line through $M$, which makes $\measuredangle \theta$ with the $x$-axis, to intersect the given semicircle at $Q$. Draw the line $OT$ through $O$ parallel to $MQ$. Draw a line through $Q$ parallel to $x$-axis to meet $OT$ at $H$. Drop a perpendicular from $H$ to $MQ$ to meet it at $G$. Now, $G$ is the center of the sought semicircle and $GQ$ is its radius. Draw the semicircle $EGF$ to complete the construction.
In the second scenario illustrated in $\mathrm{Fig.\space 3}$, $r_1$, the radius of the sought semicircle, is given. Extend the $y$-axis up to $J$ so that $AJ = r_1$. Draw a circular arc with $M$ as the center and $MJ$ as the radius. Construct a right angle triangle $OKB$ such that $KB = r_1$ and $OB = R$. Draw a second circular arc with $O$ as the center and $OK$ as the radius to intersect the first arc at $G$, which happens to be the center of the sought semicircle having the given radius $r_1$.

