# line tangent to 2 circles....

• Nov 8th 2006, 02:42 PM
Nitz456
line tangent to 2 circles....
http://img384.imageshack.us/img384/2219/glooberqu7.jpg

I've tried literally everything I have in my knowledge to try to get this problem. I am to find the tangent linewith a positive slope to these two circles (the one on the left is x^2+y^2=1, the one on the right is y^2+(x-3)^2=4), and find the two points which the line hits these circles. Seriously, any help at all, would be the most fantastic thing I'd ever read. Thanks.
• Nov 8th 2006, 03:27 PM
topsquark
Quote:

Originally Posted by Nitz456
http://img384.imageshack.us/img384/2219/glooberqu7.jpg

I've tried literally everything I have in my knowledge to try to get this problem. I am to find the tangent linewith a positive slope to these two circles (the one on the left is x^2+y^2=1, the one on the right is y^2+(x-3)^2=4), and find the two points which the line hits these circles. Seriously, any help at all, would be the most fantastic thing I'd ever read. Thanks.

Well, to start with, it's a mess. :) But I can at least maybe get you started.

Let the tangent line be of the form: $y = mx + b$

Let the point where this line touches the rightmost circle be $(x_1, y_1)$. Right off we know that
$y_1 = mx_1 + b$

Since we know this line is going to touch the top of the circle, we also know that
$y_1 = + \sqrt{1 - x_1^2}$

Now, the slope of the line tangent to the circle will be equal to the derivative of the circle function at that point. So:
$y = + \sqrt{1 - x^2}$

$y' = \frac{1}{2} \cdot \frac{1}{\sqrt{1-x^2}} \cdot (-2x) = \frac{-x}{\sqrt{1-x^2}}$

At $x = x_1$ this must be the slope of the tangent line. So:
$m = \frac{-x_1}{\sqrt{1-x_1^2}}$

We may do a similar analysis on the leftmost circle and calling the tangent point $(x_2, y_2)$ we get:
$y_2 = mx_2 + b$

$y_2 = + \sqrt{4 - (x_2 - 3)^2}$

$m = \frac{-x_2 + 3}{\sqrt{4-(x_2 - 3)^2}}$

This is 6 conditions on 6 variables $(x_1, y_1, x_2, y_2, m, b)$ so this system should be solvable.

There will be many ways to solve this system. My advice is to start with the linear equations and try to eliminate as many variables as you can using these. Also note the form of the "derivative" equations:
$m = \frac{-x_1}{\sqrt{1-x_1^2}}$

and the circle equations:

$y_1 = + \sqrt{1 - x_1^2}$

Note then, that we may use the second equation in the first:
$m = \frac{-x_1}{y_1}$

We can pull the same trick on the second circle. This should help eliminate some of the mess in the equations.

Good luck!

-Dan
• Nov 8th 2006, 05:15 PM
ThePerfectHacker
I recommend you to read how to geometrically (using Euclid's toys) you can construct a common tangent.
• Nov 8th 2006, 07:34 PM
Nitz456
alright I've decided to go a different route then yours, but thank you so much for all of that information. I have one thing stopping me from solving this problem, and that is:

I cannot find the value of Y for when the tangent line is at x=-3. I know it's zero, but I can't prove it. It's driving me insane.

I'm using this method if anyone is wondering:

Let's do this right. I had to leave and come back.

The small circle has equation y=(1-x^2)

The large circle: y=(-x^2+6x-5)

Find the derivatives, set them equal and solve for x.

-x/(1-x^2)=(3-x)/(-x^2+6x-5)

We find, with lotsa algebra, that x=-3 and y=0

This is where the line crosses the x-axis. We now have a point to work with.
• Nov 8th 2006, 08:12 PM
malaygoel
Quote:

Originally Posted by Nitz456
alright I've decided to go a different route then yours, but thank you so much for all of that information. I have one thing stopping me from solving this problem, and that is:

I cannot find the value of Y for when the tangent line is at x=-3. I know it's zero, but I can't prove it. It's driving me insane.

I'm using this method if anyone is wondering:

Let's do this right. I had to leave and come back.

The small circle has equation y=(1-x^2)

The large circle: y=(-x^2+6x-5)

Find the derivatives, set them equal and solve for x.

-x/(1-x^2)=(3-x)/(-x^2+6x-5)

There is a mistake here.
While setting them equal, you have to remember that the slope you have found are at two different points.
$\frac{-x_1}{\sqrt{1-x_1^2}}=\frac{3-x_2}{\sqrt{-x_2^2+6x-5}}$
where $x_1$and $x_2$ are the points where the common tangent meet the first and second circles respectively.
Keep Smilimg
Malay
• Nov 9th 2006, 04:22 AM
earboth
Quote:

Originally Posted by Nitz456
http://img384.imageshack.us/img384/2219/glooberqu7.jpg

I've tried literally everything ... to find the tangent linewith ... to these two circles ...

Hello, Nitz,

I'll show you how you can construct this tangent geometrically. (See attachment)

1. Draw the circles c1 and c2 (thick black line)
2. Draw the auxiliar circle (blue line) with radius = r2 - r1
3. Construct the Thales circle between the centres of c1 an c2 (green line)
4. Now you can construct the tangent ta from the centre of c1 to the auxiliar circle.
5. Move ta out until it touches c1 and c2.

You probably have noticed that there is another common tangent which is perpendicular to the x-axis at x = 1.

EB