Geometry problem: finding coordinates on outer circle using extrusion from centre through point of inner circle

Question

My problem involves an inner circle (centre $I$) with a displaced placement within an outer circle (centre $O$). I have a point placed somewhere on the circumference of the inner circle ($P$). I wish to extrude from point $P$ orthogonally until it reaches and intersects the outer circle at position $Q$, so I need to calculate the position of $Q$.

The radius of the outer circle ($r_1$), inner circle ($r_2$) and displacement of inner circle ($r_3$) are known.

Any help would be greatly appreciated.

Some background:

I actually need to do this for multiple points on the inner circle ($P_1$, $P_2$ etc, but I have all these coordinates). What I have so far:

$x_Q = x_I + \frac{(x_P - x_I) \cdot r_1}{\sqrt{(x_P - x_I)^2 + (y_P - y_I)^2}}$

$y_Q = x_I + \frac{(y_P - y_I) \cdot r_1}{\sqrt{(x_P - x_I)^2 + (y_P - y_I)^2}}$

which gives me something like this, but the magnitude is wrong:

I then tried adding the original displacement of the inner circle,

$x_Q = x_I + \frac{(x_P - x_I) \cdot r_1}{\sqrt{(x_P - x_I)^2 + (y_P - y_I)^2}} + r_3$

$y_Q = x_I + \frac{(y_P - y_I) \cdot r_1}{\sqrt{(x_P - x_I)^2 + (y_P - y_I)^2}}$

but now the lines aren't orthogonal:

Answer 1

Hint: Let $O,I,P$ be the outer and inner circle centers, and the point on the inner cirle. Let $d(\cdot,\cdot)$ be the Euclidean distance, and $t$ a parameter to be found. Set up the equation $$d(O,\ I+t(P-I))=r$$ where $r$ is the radius of the larger "outer" circle. This may be solved for $t$, make things easier by squaring the sides first. For your purpose you only want solution with $t>0.$ Once you have a number for $t$ you can fill it in to $I+t(P-I)$ to get the desired point on the outer circle.

Added Another description: With $P,I,O$ as the points viewed as vectors, a dot product version gives the quadratic equation $$at^2+bt+c=0, \\ a=(P-I)\cdot (P-I),\\ b=2(I-O)\cdot(P-I) \\ c=(I-O) \cdot (I-O) - r^2$$

Answer 2

In order to do this, you presumably know the coordinates of $I$ and $O$ in your figure. Then, knowing the coordinates of $P$, you can compute the angle $\angle OIP$ and its sine, $\sin(\angle OIP)$.

Specifically, let $u$ be the vector from $I$ to $O$ and let $u$ be the vector from $I$ to $P$. Then

\begin{align} \cos(\angle OIP) &= \frac{u \cdot v}{\lVert u\rVert\lVert v\rVert}, \\ \sin(\angle OIP) &= \sqrt{1 - \cos^2(\angle OIP)}, \\ \angle OIP &= \arccos(\cos(\angle OIP)). \end{align}

Note that this always gives an angle between zero and $\pi$ radians, so you'll also need to keep track of which "side" of the line $IO$ the point $P$ is on so you can project it to the correct point on the outer circle and not the mirror-image point across line $IO.$

Now apply the law of sines and the fact that $\angle OIP = \angle OIQ$ in the triangle $\triangle OIQ$:

$$ \frac{\sin(\angle OIP)}{\angle OIP} = \frac{\sin(\angle IQO)}{\angle IQO}. $$

Therefore $$ \angle IQO = \frac{\sin(\angle IQO)}{\sin(\angle OIP)}\angle OIP. $$

Then $$ \angle IOQ = \pi - \angle OIP - \angle IQO. $$

From $\angle IOQ$ and the radius $r_1$ you can project $Q$ from $O$, just taking care to avoid the mirror-image location.

Answer 3

We use polar coordinates. We can ignore point P because it anyhow lies on radius IQ and that PQ is always perpendicular to inner circle centered at I. No need even to label it.

$$ IQ=r, ~~ IO =d, \text{ labeled in place of }r_3; $$

By Cosine Rule on triangle $I Q_1 O$ we have

$$r_1^2=r^2+d^2-2rd \cos(\pi-\theta),$$

$$ r^2+2rd \cos \theta +(d^2-r_1^2) =0 ~;$$

Solutions of the quadratic equation

$$ r_{1,2}=d\cos \theta\pm \sqrt { r_1^2 - (d \sin \theta)^2)} $$

The (the extended ) radius vector intersects the outer circle at two points $Q_1,Q_2$ centered at $I$ as shown above.

$$ (x_{Q_1},y_{Q_1})= r_1( \cos\theta, \sin\theta); (x_{Q_2},y_{Q_2})= r_2( \cos\theta, \sin\theta).~$$

EDIT

A polar plot and Mathematica code is also attached:

Slight label change. Red circle radii $\rho_{1,2}$ at $ Q_{1,2}$ is eccentric to circle of radius $\rho_2$ centered at I.