2
$\begingroup$

diagram of problem

Having a really hard time solving this problem. Given:

  • a circle of radius $a$
  • an ellipse with minor axis $g$ and major axis $f$
  • the ellipse is oriented so that the major axis is parallel with the vector between the circle and ellipse
  • lines which are tangent to both the ellipse and the circle while crossing between them
  • $d_1$ is the distance from the center of the circle to the crossing point
  • $d_2$ is the distance from the center of the ellipse to the crossing point
  • $d_3$ is the horizontal distance from the ellipse tangent to the crossing point
  • $d_4$ is the distance from the center of the circle to the center of the ellipse
  • $d_5$ is the horizontal distance from the center of the ellipse to the ellipse tangent
  • $b$ is the vertical distance to the ellipse tangent
  • $L$ is the distance from the crossing point to the ellipse tangent
  • I have determined the following relationships

    • $\sin(\theta_1) = \frac{a}{d_1}$
    • $\sin(\theta_2) = \frac{b}{L}$
    • $\theta_1$ = $\theta_2$
    • $L = \sqrt{d_3^2 + b^2}$
    • $d_4 = d_1 + d_2$
    • $d_2 = d_3 + d_5$
    • $\frac{b^2}{g^2} + \frac{d_5^2}{f^2} = 1$
    • $\tan(\theta_1) = \frac{b}{d_3}$

To clarify, the known values are:

  • $a$
  • $g$
  • $f$
  • $d_4$

I would like to solve for $d_2$. Any assistance would be greatly appreciated.

$\endgroup$
0
$\begingroup$

Here's a coordinate-based approach.


Set the center of the circle at the origin. Then the tangent line through $T(a \cos\phi, a \sin\phi)$ has $x$- and $y$-intercepts $a\sec\phi$ and $a\csc\phi$, respectively. (Note that the $x$-intercept is $d_1$ in your figure.) In intercept-intercept form, the line's equation is $$\frac{x}{a\sec\phi} + \frac{y}{a\csc\phi}=1 \qquad \text{or, more simply,} \qquad x \cos\phi + y\sin\phi = a$$ We can determine the intersections of the line with the ellipse $$\frac{(x-d_4)^2}{f^2}+\frac{y^2}{g^2}=1$$ by replacing $y$ with $(a - x \cos\phi)/\sin\phi$. This gives $$\begin{align}0 &= x^2 ( f^2 \cos^2\phi + g^2 \sin^2\phi ) - 2 x ( a f^2 \cos\phi + d_4 g^2 \sin^2\phi ) &(\star)\\ &\quad+ a^2 f^2 + d_4^2 g^2 \sin^2\phi - f^2 g^2 \sin^2\phi \end{align}$$ Because (for $\phi$ corresponding to a line that is tangent to the ellipse) the line should have just one intersection with the ellipse, we know that $(\star)$ must have only one root, $x$; thus, its discriminant must vanish: $$-4 f^2 g^2 \sin^2\phi \; ( ( d_4^2 - f^2 + g^2 ) \cos^2\phi - 2 a d_4 \cos\phi + a^2 - g^2 ) = 0 \quad (\star\star)$$ Presumably, $f$ and $g$ are non-zero. If $\sin\phi = 0$, then our common tangent line is vertical; this requires that the ellipse be tangent to the circle, so that $d_2 = f$. In general, we must have that the final factor of $(\star\star)$ vanishes; we can solve this quadratic for $\cos\phi$: $$\cos\phi = \frac{a d_4 \pm \sqrt{a^2 d_4^2 - (d_4^2-f^2+g^2)(a^2-g^2)}}{d_4^2-f^2+g^2} \qquad (\star\star\star)$$ Recall that $d_1 = a\sec\phi = a/\cos\phi$. From $(\star\star\star)$, we have $$\begin{align} d_1 &= \frac{a(d_4^2-f^2+g^2)}{a d_4 \pm \sqrt{a^2 d_4^2 - (d_4^2-f^2+g^2)(a^2-g^2)}} \\ &= \frac{a(d_4^2-f^2+g^2)(a d_4 \mp \sqrt{a^2 d_4^2 - (d_4^2-f^2+g^2)(a^2-g^2)}}{a^2 d_4^2 - \left(a^2 d_4^2 - (d_4^2-f^2+g^2)(a^2-g^2)\right)} \\ &= \frac{a(a d_4 \mp \sqrt{a^2 d_4^2 - (d_4^2-f^2+g^2)(a^2-g^2)}}{a^2-g^2} \end{align}$$ and then $$\begin{align} d_2 = d_4 - d_1 = \frac{-g^2 d_4 \pm a\sqrt{a^2 d_4^2 - (d_4^2-f^2+g^2)(a^2-g^2)}}{a^2-g^2} \end{align}$$

Some notes:

  • The "$\pm$" ambiguity accounts for the fact that there are four tangent lines (corresponding to two values of $\cos\phi$) common to the circle and ellipse. The "internal" pair cross between the centers; the "external" pair enclose the circle and ellipse. (The external pair may be parallel. Specifically, the external pair is parallel when $g=a$, which is why the formula for $d_2$ seems averse to this situation: if the lines are parallel, $d_2$ is infinite.)
  • When $d_4 = a + f$ (the circle and ellipse are tangent, and the tangent line is vertical), the above formula (with $\pm=+$) reduces to $d_2 = f$. So, the formula incorporates the $\sin\phi = 0$ case from before.
  • When the ellipse is a circle of radius $b$ (not the $b$ in your figure) ---that is, when $f = g = b$--- then the formula (with $\pm=+$) reduces to $d_2 = \frac{d_4 b}{a+b}$. (Correspondingly, $d_1 = \frac{d_4 a}{a+b}$.) It's left to the reader to verify that this is consistent with the figure.
  • When the ellipse is a circle congruent to the given circle ---that is, when $f = g = a$--- the formula in the previous bullet reduces to $d_2 = d_4/2$. (It's problematic to directly substitute $f=a$ and $g=a$ into the $d_2$ formula.) This is clearly consistent with the figure.
$\endgroup$
1
$\begingroup$

Assume the crossing point of the tangents at the origin, the center of the circle at $(-p,0)$, and the center of the ellipse at $(q,0)$.

A line $y=mx$ is tangent to the circle $(x+p)^2+y^2=a^2$ when the intersection of the two results in a quadratic equation with discriminant $0$. The computation leads to $a^2(1+m^2)- m^2 p^2=0$, or $$m^2={a^2\over p^2- a^2}\ .\tag{1}$$ Similarly, the line $y=mx$ is tangent to the ellipse $g^2(x-q)^2+f^2y^2=f^2g^2$ when the intersection of the two results in a quadratic equation with discriminant $0$. The computation leads to $(g^2+m^2f^2)- m^2 q^2 =0$, or $$m^2={g^2\over q^2-f^2}\ .\tag{2}$$ From $(1)$ and $(2)$ it follows that $$a^2(q^2-f^2)=g^2(p^2-a^2)\ .$$ Together with $p+q=d_4$ this allows to compute $p$ $(=d_1$) and $q$ ($=d_2)$.

$\endgroup$
  • $\begingroup$ Thanks Christian. Your answer was consistent with Blue - the only reason I picked his as the answer was for the additional insight on the solution permutations. The different perspective is useful though and I really appreciate your effort. $\endgroup$ – rccola Jan 21 '14 at 16:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.