Compass-and-Straightedge Construction 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.
 A: 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 $l.$

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 $M.$

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 $A_1PQ, A_2PQ.$   
A: *

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

*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. 

*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"). 

*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. 
A: 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.
