Compass-and-Straightedge Construction

Question

I stumbled upon this question in math class, and I got stuck. The Question: You're are given a circle, and two points. How do you construct a circle that goes through the two points and is tangent to the the given circle? Thank you, please reply.

Answer 1

1. You can't do this in general if one point is inside the circle and the other's outside.

2. Assuming both are outside, there are generally two solutions, a "small" and a "large" one. But if the line $PQ$ between the two points, $P$ and $Q$, contains the center $C$ of the circle, then the two solution circles are the same size.

3. If both points are on the circle, then there is no solution (or the circle itself is the solution -- I suppose it depends on your definition of "tangent").

4. If one point, $P$ is on the circle and the other, $Q$ is outside, you draw a ray from the circle-center $C$ through $P$, and the perpendicular bisector of $PQ$; where these intersect is the center $D$ of the new circle, with radius $DP$.

Case 2 is the tough one, though. I just figured that helping settle the easy (and impossible) cases might be useful. I don't, offhand, have a solution for case 2. But @A.P.'s remark that it's a special case of Appolonius's problem pretty much takes care of it.

Answer 2

i can do in the case of two points that are both outside or both inside the given circle. i will use inversion on circle.

in fact we can take the circle to be a line and the points on the same side of the line. all you have to do is invert on a circle that goes through the center of the given circle.

let the given line be $l$ and the two pints $P$ and $Q$ on the same side of $l.$

first we want to get rid of the easier case where the line $PQ$ is parallel to the line $l.$

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here are the steps:

(a) perpendicular bisector of $PQ$ cut $l$ at $M.$

(b) perpendicular bisectors of $MP, MQ$ cut at $O.$

(c) $O$ is the center of the required circle through $P, Q$ and touching $l$ at $M.$

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here are steps to construct the two circles through $P, Q$ and touching $l:$

(a) line $PQ$ cuts $l$ at of $M.$

(b) draw a circle $C$ that has $PQ$ for diameter.

(c) draw two tangents from $M$ to $C.$ let it touch at points $T_1, T_2.$

(d) draw a circle $\omega$ with center $M$ and radius $MT_1 = MT_2.$

(e) let $\omega$ cut line $l$ at $A_1, A_2$

(f) the required circles are circles through $A_1PQ, A_2PQ.$

Answer 3

You can reduce this to a problem in analytic geometry, but the answer is pretty brutal. If you set up your coordinates so the two points are at $(0,d)$ and $(0,-d)$, the circle's center is at $(a,b)$ and the circle's radius is $r$, then you are looking for a point on the $x$-axis, let's say $(x,0)$, such that the distance from $(x,0)$ to $(a,b)$ is $r$ plus the distance from $(x,0)$ to $(0,d)$. This gives the equation

$$\sqrt{(x-a)^2+d^2}=r+\sqrt{x^2+d^2}$$

Solving this for $x$ gives a result that looks horrendous but can be constructed with compass and straightedge. Here is the simplest way I have found so far to calculate $x$:

$$x=\frac {a(a^2+b^2-d^2-r^2)\pm r\sqrt{4a^2d^2+(a^2+b^2)^2+(d^2-r^2)^2-2(a^2+b^2)(d^2+r^2)}} {2(a^2-r^2)}$$

Several shortcuts present themselves, but this will still be a monstrous construction.