Ratio of Circle Radii that Share tangent lines

• Oct 4th 2010, 05:10 PM
arcanine
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. $\displaystyle P_0T_{11}$ and $\displaystyle P_0T_{12}$ are tangent to BOTH circles. $\displaystyle P_1P_2$ is tangent only to circle 2 (the larger circle).

Conceptually, I think this is easy. The center of circle 2 ($\displaystyle C_2$) lies on $\displaystyle P_0P_2$, and the radius is such that it is tangent to both $\displaystyle C_1$ and $\displaystyle C_2$. 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 $\displaystyle C_2$ 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 $\displaystyle d(\textbf{x}_0,\textbf{x}_1)$ is the distance formula between points $\displaystyle \textbf{x}_0$ and $\displaystyle \textbf{x}_1$:
$\displaystyle d(P_0,T_2_1)=d(P_0,T_2_2)$
$\displaystyle d(P_1,T_2_2)=d(P_1,T_3)$
$\displaystyle d(P_0,T_2_2)+d(T_2_2,P_1)=d(P_0,P_1)$

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

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:
$\displaystyle d=\dfrac{\mid(\textbf{x}_0-\textbf{x}_1)\times(\textbf{x}_0-\textbf{x}_2)\mid}{\mid\textbf{x}_2-\textbf{x}_1\mid}$

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

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

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.
• Oct 4th 2010, 08:00 PM
sa-ri-ga-ma
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

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

find r2. T11 is the radius of C1. Now find the ratio of r1 and r2.
• Oct 5th 2010, 06:25 AM
arcanine
I'm not sure I understand your post
Quote:

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, $\displaystyle P_1$ and $\displaystyle P_2$ have the same y-coordinates, so,
$\displaystyle y=c=y_{P_1}$

Quote:

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

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

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 $\displaystyle C_1$, and if it did, it would only be a coincidence.
• Oct 5th 2010, 06:47 AM
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?

Then how can be their y-coordinates are the same? In the picture they are not the same.
• Oct 5th 2010, 10:55 AM
arcanine
Quote:

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 $\displaystyle P_1$ and $\displaystyle P_2$ 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 $\displaystyle C_2$ would become a known point.
• Oct 5th 2010, 12:22 PM
icemanfan
Assuming $\displaystyle P_1P_2$ is horizontal, then $\displaystyle C_2T_3$ is vertical, and the equation for the line is $\displaystyle x = a$ for some value of $\displaystyle a$. Now the line $\displaystyle T_{22}C_2$ is perpendicular to $\displaystyle P_0P_1$ and therefore you know its slope $\displaystyle m_2$.

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

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

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

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

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

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

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

Now the distances $\displaystyle T_{22}C_2$ and $\displaystyle T_3C_2$ must be equal, therefore $\displaystyle \sqrt{(a - x_0)^2 + (m_1a + b_1 - y_0)^2} = m_1a + b_1 - b$. 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.
• Oct 5th 2010, 10:06 PM
sa-ri-ga-ma
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.
• Oct 6th 2010, 07:07 AM
arcanine
Quote:

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.