1. Ratio of Circle Radii that Share tangent lines

I don't know how to explain the arrangement, so I've attached an image. Forgive my notation, I'm not used to writing math problems.

The points that are black are known. The points that are red are unknown. $P_0T_{11}$ and $P_0T_{12}$ are tangent to BOTH circles. $P_1P_2$ is tangent only to circle 2 (the larger circle).

Conceptually, I think this is easy. The center of circle 2 ( $C_2$) lies on $P_0P_2$, and the radius is such that it is tangent to both $C_1$ and $C_2$. Without any other conditions, there would be an infinite number of solutions. However, there is an additional line that bounds circle 2. So, unless I am mistaken, there is now only one solution.

However, I don't know what I'm doing wrong. I've tried using similar triangles, but because neither $C_2$ nor the radius for circle 2 is known, I haven't been able to determine the ratio of the two triangles. In fact, everything I've tried seems to require a system of equations.

So I tried the following, where $d(\textbf{x}_0,\textbf{x}_1)$ is the distance formula between points $\textbf{x}_0$ and $\textbf{x}_1$:
$d(P_0,T_2_1)=d(P_0,T_2_2)$
$d(P_1,T_2_2)=d(P_1,T_3)$
$d(P_0,T_2_2)+d(T_2_2,P_1)=d(P_0,P_1)$

However, when I write the distance formulas for three dimensions, I end up with nine unknowns; three for each of the following points, $T_2_1$, $T_2_2$, and $T_3$.

So, I tried an additional three equations based on the distance of the center of circle 2 to each of the bounding lines using the following formula:
$d=\dfrac{\mid(\textbf{x}_0-\textbf{x}_1)\times(\textbf{x}_0-\textbf{x}_2)\mid}{\mid\textbf{x}_2-\textbf{x}_1\mid}$

I needed three more equations, so I tried parameterizing the equation for $P_0P_1$, yielding:
$x_{T_{22}}=x_{P_0}+t(x_{P_1}-x_{P_0})$
$y_{T_{22}}=y_{P_0}+t(y_{P_1}-y_{P_0})$
$z_{T_{22}}=z_{P_0}+t(z_{P_1}-z_{P_0})$

where $t=\dfrac{d(P_0,T_{22})}{d(P_0,P_1)}$

However, this led to a binomial equation in x, y and z. So, instead I tried using the equation for each of the three bounding lines, since the respective tangent point is a solution to the corresponding line equation. Unfortunately, I'm not sure if I'm not solving the system of equations correctly, or if my equations are not independent from each other.

Again, I feel like there should be an elegant solution to this. Perhaps something similar to finding the tangent points of a circle inscribed inside a triangle.

Any help would be greatly appreciated.

2. Hi arcanine, welcome to MHF.

P1 and P2 are known points.

Find the equation of the tangent P1P2.

Write it in the form y = mx + c.

Condition for line to be a tangent to a circle C2 is

$c^2 = r_{2}^2(1 + m^2)$

find r2. T11 is the radius of C1. Now find the ratio of r1 and r2.

3. I'm not sure I understand your post

Originally Posted by sa-ri-ga-ma
Find the equation of the tangent P1P2.

Write it in the form y = mx + c.
In this case, $P_1$ and $P_2$ have the same y-coordinates, so,
$y=c=y_{P_1}$

Condition for line to be a tangent to a circle C2 is

$c^2 = r_{2}^2(1 + m^2)$

find r2. T11 is the radius of C1. Now find the ratio of r1 and r2.
When using the above equation, I get a result that is not the radius of circle 2 (See Attached Image). Also note that Circle 3 does not pass through $C_1$, and if it did, it would only be a coincidence.

4. In the problem you have mensioned that P1 and P2 are known points. (The points that are black are known) Is it so?

Then how can be their y-coordinates are the same? In the picture they are not the same.

5. Originally Posted by sa-ri-ga-ma
In the problem you have mensioned that P1 and P2 are known points. (The points that are black are known) Is it so?
Yes, it is so. Perhaps my assumptions are not standard, but I was assuming that the y-axis was a vertical line on the screen, and the x-axis was a horizontal line on the screen, such that they are perpendicular to eachother. Under that assumption, the y-coords for $P_1$ and $P_2$ are the same.

However, I'm not too concerned with their y-coordinates. I'm more interested in how you derived the equation for a line tangent to a circle, because it is not giving the correct radius of the circle, as mentioned in post #3. An important note is that the origin is at an arbitrary location.

Without actually writing things out, it appears that the equation you provided for a line tangent to a circle assumes that the circle's center is at the origin. If this were the case, I wouldn't have any problems, because $C_2$ would become a known point.

6. Assuming $P_1P_2$ is horizontal, then $C_2T_3$ is vertical, and the equation for the line is $x = a$ for some value of $a$. Now the line $T_{22}C_2$ is perpendicular to $P_0P_1$ and therefore you know its slope $m_2$.

So since $P_0P_2$, $C_2T_3$ and $T_{22}C_2$ all intersect at the same point, you have:

$T_{22}C_2: y = m_2x + b_2$
$P_0P_2: y = m_1x + b_1$
$C_2T_3: x = a$
$P_1P_2: y = b$
$P_0P_1: y = m_3x + b_3$

Therefore the point of intersection can be written:
$(a, m_1a + b_1)$

where $a$ and $b_2$ are the only unknowns.

Now, $m_1a + b_1 = m_2a + b_2$, which yields
$b_2 = (m_1 - m_2)a + b_1$

Continuing, we find the point of intersection of $P_0P_1$ and $T_{22}C_2$ in terms of a:

$m_2x + b_2 = m_3x + b_3$
$m_2x + (m_1 - m_2)a + b_1 = m_3x + b_3$
$(m_2 - m_3)x = b_3 - (m_1 - m_2)a$
$x_0 = \frac{b_3 - (m_1 - m_2)a}{m_2 - m_3}$
$y_0 = \frac{m_3(b_3 - (m_1 - m_2)a)}{m_2 - m_3} + b_3$

Now the distances $T_{22}C_2$ and $T_3C_2$ must be equal, therefore $\sqrt{(a - x_0)^2 + (m_1a + b_1 - y_0)^2} = m_1a + b_1 - b$. Since everything is in terms of a or known constants, solve the equation for a and then you will know the location of every unknown point.

7. As you have guessed, I have assumed that Po is the origin. When you say that some points are known, you must mention the poisition of the origin.

8. Originally Posted by sa-ri-ga-ma
When you say that some points are known, you must mention the poisition of the origin.
I apologize, I'm new to describing math problems.

I've looked at icemanfan's solution, and I believe it will work. I haven't crunched through the math to be sure, but at first glance it looks good.

Thanks everyone for the help.