Comments on How can we prove that a point that follows another point has the same trajectory given a contant angle and ratio?
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How can we prove that a point that follows another point has the same trajectory given a contant angle and ratio?
So suppose that we have a point called $P$ that moves along the circle $O_1$. The parametric equations of $P$ are given by:$$\begin{aligned}x(t) &= \cos(t) + 5 \\y(t) &= \sin(t) + 3 \\&\text{Or}\\\overrightarrow{P(t)} &= \begin{Bmatrix}\cos(t) + 5 \\\sin(t) + 3\end{Bmatrix}\end{aligned}$$ This point $P$ is connected to the point $A(0,0)$. Another line segment from $A$ that is connect to a point is called $Q$. The angle between the two line segments is $\alpha$, and the ratio between the lengths of the two segments is a constant $k$. We want to determine the trajectory of point $Q$. It seems obvious that the trajectory of $Q$ is a circle, but I found out that it's quite difficult to prove it. Because I couldn't see how to use the dot product correctly.
Here is an image of the situation I described (one of the possible situations):
I attempted to use the dot product as follows:$$\cos(\alpha)=\frac{\overrightarrow{P(t)}\cdot\overrightarrow{Q(t)}}{|\overrightarrow{P(t)}|\cdot|\overrightarrow{Q(t)|}}$$ However, this approach did not help me isolate the term involving $\overrightarrow{Q(t)}$. I could have proven that the trajectory of $Q$ is a circle if $\alpha$ is $90$ degrees, but not for a general angle $\alpha$. Is there an extra formula that I need to use or did I need to use another approach?
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The dot product cannot be sufficient because even making substitutions to minimise the occurrences of $Q$ we get $$\cos(\alpha)=\frac{\overrightarrow{P(t)}\cdot\overrightarrow{Q(t)}}{k|\overrightarrow{P(t)}|^2}$$ there are two solutions in the plane.
I personally would tackle this by complex numbers, but to keep it more geometric we can express the relationship as $$Q(t) = k \begin{pmatrix} \cos(\alpha) & \sin(\alpha) \\ -\sin(\alpha) & \cos(\alpha) \end{pmatrix} P(t)$$
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