Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose we are given a ray $\rho_a$ beginning at point $a$ and a ray $\rho_b$ beginning at point $b$. I want to find a circle $C_a$ tangent to $\rho_a$ at point $a$ and another circle $C_b$ tangent to $\rho_b$ at point $b$ such that there exists some point $c$ and a line $\ell$ passing through $c$ such that $C_a$ and $C_b$ intersect at $c$ tangent to $\ell$.

Note that if we insisted on a particular point $c$ and line $\ell$ given ahead of time, then each circle would be over-constrained. However, since we don't insist on a particular point $c$, I think that this should be possible.

rays


Edit: Here are two possible solutions (approximate, b/c it's hard to draw)

share|improve this question
add comment

2 Answers 2

I'll assume $b$ is not on $\rho_a$. Start with a circle $C_a$ through $a$ tangent to $\rho_a$ and with $b$ is inside it (any large enough circle through $a$ tangent to $\rho_a$ and on the same side as $b$ will do). Say $d_a$ is its centre and $r_a$ its radius. Let $d_b$ be a point on the line through $b$ perpendicular to $\rho_b$ such that $\|d_b - d_a\| + \|d_b - b\| = r_a$. Then the circle $C_b$ centred at $d_b$ with radius $\|d_b - b\|$ passes through $b$, and $C_a$ and $C_b$ are tangent at the point $c$ where the line through $d_a$ and $d_b$ intersects $C_a$ on the other side of $d_b$.

EDIT: Here's the geometric construction. Let $L$ be the line through $b$ perpendicular to $\rho_b$. Let $p$ be on this line at distance $r_a$ from $b$, on the same side of $\rho_b$ as $d_a$. Then the perpendicular bisector of the line segment from $p$ to $d_a$ intersects $L$ at $d_b$ which is between $b$ and $p$ (because $b$ is closer to $d_a$ than it is to $p$), and $\|d_b - d_a\| + \|d_b - b\| = \|p - d_b\| + \|d_b - b\| = r_a$.

enter image description here

share|improve this answer
    
Given your answer, I can find the circles algebraically. Is there a construction e.g. with compass, ruler, protractor. (I want to draw it by hand). –  Joe Sep 25 '12 at 1:11
add comment

Your system is, at the moment, underdetermined. Through $a$, parallel to $\rho_a$, there is a continuous one-parameter family of circles with centers on the line perpendicular to $\rho_a$ and passing through $a$. This family is called a "pencil" of circles. Likewise, there is a pencil associated to the line $\rho_b^\perp$.

For any circle in the pencil of $a$, there are two circles in the pencil of $b$ which are tangent to it, although I do not immediately see a compass/ruler/protractor algorithm to draw them. The argument for existence uses some version of the intermediate value theorem.

The family of tangency points seems like it should follow some algebraic curve, perhaps a hyperbola? The nature of that curve would be interesting to study.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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