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(Essentially this question, but in 3 dimensions.)

In 3D space, points $a$ and $b$ are known. In addition, a unit vector $\hat{t}$ is known. Assume $b-a$ and $\hat{t}$ are not parallel.

The goal: determine the center point of a circle which passes through both $a$ and $b$ and is tangent to $\hat{t}$ at point $a$.

(The circle will then lie on the plane spanned by $b-a$ and $\hat{t}$.)

My attempt:

Let $c$ be the center of the circle (the goal).

Get a vector in the direction of the normal of the plane $n$:

$$n=(b-a)\times\hat{t}$$

Find three equations that can be used to solve for $[c_x, c_y, c_z]$.

First, using the fact that $(a-c)$ and $\hat{t}$ must be perpendicular, $(a-c)\cdot\hat{t}=0$, or

$t\cdot c = t \cdot a \tag{1}$

Second, $(b-c)$ must be perpendicular to $n$:

$n \cdot c = n \cdot b \tag{2}$

Third, $c$ must lie on the perpendicular bisecting plane of $a$ and $b$:

$(b-a)\cdot c = (\frac{b-a}{2})\cdot (b-a) \tag{3}$

This gives the equation

$$ \begin{bmatrix} t_x & t_y & t_z \\ n_x & n_y & n_z \\ b_x-a_x & b_y-a_y & b_z-a_z \end{bmatrix} \begin{bmatrix} c_x \\ c_y \\ c_z \end{bmatrix} = \begin{bmatrix} t \cdot a \\ n \cdot b \\ (\frac{b-a}{2})\cdot (b-a) \end{bmatrix} $$

However, solving that (by plugging in $a$, $b$, and $\hat{t}$ and doing it on a computer) is giving me wildly off results. $c$ is in the correct plane, but that's about it.

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  • $\begingroup$ do you mean with $c$ the center of the circle? how are ($a-c$) and $\hat{t}$ prependicular ? $\endgroup$ Commented Sep 2, 2020 at 17:42

3 Answers 3

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With $p = (x,y,z)$ the circle's center is defined by the intersection of the following three planes

$$ \cases{ \Pi_1\to (p-a)\cdot \vec t = 0\\ \Pi_2\to (p-a)\cdot\left((b-a)\times \vec t\right)=0\\ \Pi_3\to (p-\frac 12(a+b))\cdot(b-a) = 0 } $$

giving $c$. Now, the circle can be parametrized as

$$ p = c + r\left(\vec e_1\cos\theta+\vec e_2\sin\theta\right) $$

with

$$ \cases{\vec e_1 = \vec t\\ \vec u = (b-a)-\left((b-a)\cdot \vec t\right)\vec t\\ \vec e_2 = \frac{\vec u}{|\vec u|}\\ r = |a-c| } $$

Included a plot when

$$ \cases{ a = (1,1,1)\\ b = (2,3,-2)\\ \vec t = (\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}) } $$

enter image description here

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Equation (3) should be: $$(b-a)\cdot (c-a)=\Big(\frac{b-a}{2}\Big)\cdot(b-a)\tag 3$$

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  • $\begingroup$ This should be a comment, not an answer. $\endgroup$ Commented Sep 3, 2020 at 12:54
  • $\begingroup$ It's the answer to the question: "what is wrong?" @TobyMak. Fix it and OP has what they want. The other answers don't actually address that. $\endgroup$ Commented Sep 3, 2020 at 13:48
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    $\begingroup$ Oh you're right that I got that equation wrong, but could you understand the logic behind this one? I don't understand what it's saying. EDIT oh nevermind I see it, that's a silly mistake by me! $\endgroup$ Commented Sep 3, 2020 at 16:26
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First $\vec{a}$ and $\vec{b}$ must lie in the plane you mentioned, but such that both are on the same side of $\hat{t}$. Set $$ \vec{m} = \vec{b} - \vec{a} $$ Second, form the vector $$ \hat{n} = \frac{\hat{t} \times (\hat{t} \times \hat{m})}{||\hat{t} \times(\hat{t} \times \hat{m})||} = \hat{t} \times \frac{\hat{t} \times \hat{m}}{||\hat{t} \times \hat{m}||} $$ where $\hat{m}=\vec{m}/||\vec{m}||$. The vector $\hat{n}$ points to the center $\vec{c}$ and is normalized. Now dot it with $\hat{m}$ you get $$ \hat{n} \cdot \hat{m} = \cos \alpha $$ where $0 \leq \alpha \leq \frac{\pi}{2}$ is the angle between the $\hat{m}$ and $\hat{n}$, and hence from $\hat{b}$ to the diameter.

Now since $\vec{m}$ is a chord in the circle, it forms a right triangle with the diameter, which forms the hypotenuse. It follows that $$ 2R = \frac{||\vec{m}||}{\cos \alpha} $$ and finally $$ \vec{c} = \vec{a} + R\ \hat{n} = \vec{a} + \frac{||\vec{m}||}{2\cos \alpha}\ \hat{n} $$

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  • $\begingroup$ Thanks for your response. I'm sorry though, as I did not include a critical part of the question. $t$ is the tangent of the circle at point $a$. I will edit the question. $\endgroup$ Commented Sep 2, 2020 at 19:13
  • $\begingroup$ So you mean that $\hat{t}$ has its start at point $a$ ? $\endgroup$ Commented Sep 2, 2020 at 19:16
  • $\begingroup$ In that case just substract $a$ from $b$, that is instead of $\vec{b}$ set everywhere $\vec{b} - \vec{a}$ and then add $\vec{a}$ to $\vec{c}$ at the end $\endgroup$ Commented Sep 2, 2020 at 19:21

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