Locus of Point P Consider a circle $\circ$ of diameter $AB$ and a constant point $C$ on $AB$. 
Consider also a random point $Q$ on $\circ$. On $QC$ (but outside of those two points) we take a point $P$ such that $\frac{AC}{CB}=\frac{QC}{CP}$.
Find the locus of point $P$. 
It's a probably a circle but haven't figured out the way of proving it yet.
 A: 
By Side-Angle-Side Similarity,
$$\frac{|AC|}{|CB|} = \frac{|QC|}{|CP|} = k \quad\implies\quad \triangle ACQ \sim \triangle BCP \quad\implies\quad \frac{|AQ|}{|BP|} = k \quad (\star)$$
for a fixed $k$. Letting $Q^\prime$ be the point where $BP$ meets the original circle, we observe that $\square AQBQ^\prime$ is a rectangle. (Why?) Therefore, $AQ \cong BQ^\prime$, and $(\star)$ tells us
$$|BP| = \frac{1}{k}|BQ^\prime|$$
Thus, the locus of $P$ is a dilation in $B$ of the locus (namely, the original circle) of $Q^\prime$.

Alternatively, let $D$ be the location of $P$ when $Q$ coincides with $B$. Then
$$\frac{|AC|}{|CB|} = \frac{|QC|}{|CP|} = \frac{|BC|}{|CD|}$$
With the help of a couple of pairs of similar triangles, one observes that $\angle AQB \cong \angle BPD$. By Thales' Theorem, $\angle AQB$ (being inscribed in a semicircle with diameter $AB$) is a right angle; by the Theorem's converse, the now-necessarily-right $\angle BPD$ must be inscribed in a semi-circle with diameter $BD$. The locus of $P$, therefore, is the full-circle with diameter $BD$.
A: I will take an algebraic approach to this question. Let $S$ be the unit circle in $\mathbb{R}^2$, with $A = (-1,0)$ and $B = (1,0)$. Let $C = (c,0)$ with $-1 < c < 1$. Then $$\frac{AC}{CB} = \frac{1 + c}{1 - c}.$$
Let $Q = (p,q)$ be a point on $S$. The line segment $CQ$ is given by $f_Q:[0,1] \rightarrow \mathbb{R}^2$ with
$$f_Q(t) = \left((1-t)c + tp,tq\right).$$
Note that one should have
$$CP = \frac{1-c}{1+c}CQ.$$
Since $f_Q$ is linear, it therefore holds that $f_Q(\frac{1-c}{1+c}) = P$. We get a map $g:S \rightarrow \mathbb{R}^2$ given by $Q \mapsto f_Q(\frac{1-c}{1+c})$. One easily checks that the locus $L = g(S)$ of points $P$ is now
$$L = \left\{ (x,y) \in \mathbb{R}^2 \mid \left(x - \left(1 - \frac{1-c}{1+c}\right)\right)^2 + y^2 = \left(\frac{1-c}{1+c}\right)^2\right\},$$
a circle with radius $\frac{1-c}{1+c}$ and center $(1 - \frac{1-c}{1+c}, 0)$. In fact, the map $g$ can be seen to be the composition of multiplication by $\frac{1-c}{1+c}$ and a translation.
