Explain how to construct circle internally tangent to a larger circle, and tangent to a point on a chord of the larger circle?

Question

This is a re-submission of a question by @John Glenn which did not get answered sufficiently. I'll add a few restrictions, to further define the problem:

$R$ = radius of larger circle

$r$ = radius of smaller circle

$\overline{AB}$ = Secant line, which intersects the concentric center with radius: $(R - 2r)$

$P_A$, $P_B$ = Points describing the secant line $(x_A,y_A),(x_B,y_B)$ are known.

$P_1$ = Point describing the center of the leftward smaller circle

$P_2$ = Point describing the center of the rightward smaller circle

Image to fully describe the situation at hand.

I am interested in determining the location of both described inner circles. Any help and explanation of the algebra involved would be greatly appreciated. I have tried to set up the system of equations to solve for either circle center and get lost in the math.

Equation for the smaller circle(s):

$(x-x_p)^2 + (y-y_p)^2 = r^2$

Equation for the large circle (centered at the origin):

$x^2 + y^2 = r^2$

Equation for the tangent line:

$y = mx + b$ where

$m = $$(y_B - y_A) \over (x_B - x_A)$, $b = mx_A-y_A$

Answer 1

The centers of the small circles lie on a circle of radius $(R - r)$, and hence can be expressed as

$C = (R - r) (\cos \phi, \sin \phi)$

The chord is specified by the known points $A$ and $B$ so it has the parametric equation:

$q(t) = v_0 + t v_1 $

where $v_0 = A$ and $v_1 = (B - A)$ and $t \in [0, 1]$

Since the distance between the center $C$ and the chord is $r$, we can find an orthogonal vector to $v_1$, let's call it $v_2$, such that

$v_2 \cdot v_1 = 0$

It is trivial to construct $v_2$ from $v_1$, namely,

$v_2 = (v_{1y} , -v_{1x} ) = (B_y - A_y, A_x - B_x )$

Now, we want to solve the equation:

$ (C - A) \cdot v_2 = \pm r | v_2 | $

Plugging in $C$, $A$ and $v_2$, the above equation becomes:

$ ((R - r) \cos \phi - A_x) (B_y - A_y) + ((R-r) \sin \phi - A_y) (A_x - B_x) = \pm r L $

where $L = \sqrt{ (A_x - B_x)^2 + (A_y - B_y)^2 }$ is equal to the length of the chord $AB$.

This equation has four solutions, with two solutions corresponding to the case when the right hand side is taken with $(+r)$, and another two solutions corresponding to the case when the right hand side is taken with $(-r)$

Answer 2

Let the center $O \, (x_0, y_0)$ be the center of the big circle. The determinant $$S = \det\left( \begin{bmatrix}x_A & y_A & 1\\ x_B & y_B & 1\\ x_O & y_O & 1 \end{bmatrix}\right)$$ is twice the signed area of the triangle $P_A P_B O$ and $$|P_B - P_A| = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2 \,}$$ is the distance between the points $P_A$ and $P_B$. Then, the distance from $O$ to the secant line determined by the points $P_A$ and $P_B$ is $$h = \frac{|S|}{|P_B - P_A|}$$ Let $l$ be the line parallel to the secant line $P_A \, P_B$ that passes through the center $O$. Since the small circle is tangent to the big circle and the secant line $P_A \, P_B$, the distance of the center $X$ of the small circle to $P_A \, P_B$ is $r$. Hence, the distance of the center $X$ to the parallel line $l$ is $|r\pm h|$. Moreover, the distance between $X$ and $O$ is $R - r$. Hence, we can write the position of $X$ as follows $$X \, = \, O\, +\, \sqrt{(R-r)^2 - (r \pm h)^2\,} \, \frac{P_B - P_A}{|P_B - P_B|} \, + \, |r \pm h| \, \frac{(P_B - P_A)^{\perp}}{|P_B - P_B|}$$ or component-wise

\begin{align} \begin{bmatrix}x \\ y\end{bmatrix} \, =\, \begin{bmatrix}x_O \\ y_O \end{bmatrix} \, + \, \frac{\sqrt{(R-r)^2 - (r \pm h)^2\,}}{|P_B - P_B|} \begin{bmatrix}x_B - x_A \\ y_B - y_A \end{bmatrix} \, + \, \frac{|r \pm h|}{|P_B - P_B|} \begin{bmatrix}- y_B + y_A \\ x_B - x_A \end{bmatrix} \end{align}

Answer 3

We give below the geometrical construction, which gives you the required circles. By the way, this is nothing but a pictorial version of the answer posted by @GeometryLover, which describes the same solution using trigonometry and algebra.

Let us denote the center of the outer and inner circles as $O$. We draw a perpendicular to the given secant line $AB$ at an arbitrary point on it, say $C$. On this perpendicular, we mark points $E$ and $G$, such that $EC = CG = r$. Next, we draw two lines $ED$ and $GF$, both of which are parallel to $AB$. Finally, we draw the circle having its center at $O$, the radius of which is equal to $R-r$. This circle cuts the line $ED$ at $P_1$ and $P_3$ and line $GF$ at $P_2$ and $P_4$ respectively.

As shown in the figure, the two circles of radius $r$ drawn with their respective centers at $P_1$, $P_2$, $P_3$, and $P_4$ touch the secant line $AB$ while externally tangential to the inner circle and internally tangential to the outer circle as well.

Using coordinate geometry and trigonometry, it can be shown that $$P_1N=\left(R-r\right)\sin\left(\omega\right)\quad\text{and}\quad NO=\left(R-r\right)\cos\left(\omega\right)\quad\text{where} $$

$$\omega = 180^o - \tan^{-1}\left(\frac{y_B – y_A}{x_B – x_A}\right) - \sin^{-1}\left(\frac{\Biggl| x_A – y_A\left(\frac{ x_B – x_A }{ y_B – y_A }\right)\Biggr|\sin\left[\tan^{-1}\left(\frac{y_B – y_A}{x_B – x_A}\right)\right]+r}{R-r}\right).$$