# Help Calculate Tangent Equation to Two Circles

• January 21st 2008, 05:35 PM
CaliMan982
Help Calculate Tangent Equation to Two Circles
Hi all, I am struggling to finish this problem and was seeing if anyone could help me.

There are two circles.

Circle 1 has a center at (2,4), and radius 4.
Circle 2 has a center at (14,9) and radius 9.

I need to find the equation for common tangent line of these two circles. The circles touch so there is just one exterior tangent line.

Any help is appreciated, Thanks.
• January 21st 2008, 05:44 PM
Jhevon
Quote:

Originally Posted by CaliMan982
Hi all, I am struggling to finish this problem and was seeing if anyone could help me.

There are two circles.

Circle 1 has a center at (2,4), and radius 4.
Circle 2 has a center at (14,9) and radius 9.

I need to find the equation for common tangent line of these two circles. The circles touch so there is just one exterior tangent line.

Any help is appreciated, Thanks.

you need to be specific what tangent line you are after. there are two external ones (the x-axis happens to be one of them) and an "internal" one. that is, the tangent line that passes through the point of intersection of the circles
• January 21st 2008, 05:49 PM
CaliMan982
Not the X-Axis Tangent
Sorry for the confusion.

I do not want the external one that is just the x-axis. I want the other external tangent like that touches the top of both circles. I hope this clarifies it.
• January 21st 2008, 05:59 PM
CaliMan982
I have uploaded a simulated picture that includes which tangent line I am trying to find. The details are not correspondning to the numbers, this is just to clarify which tangent equation I am trying to find.
• January 21st 2008, 11:12 PM
earboth
Quote:

Originally Posted by CaliMan982
Hi all, I am struggling to finish this problem and was seeing if anyone could help me.

There are two circles.

Circle 1 has a center at (2,4), and radius 4.
Circle 2 has a center at (14,9) and radius 9.

I need to find the equation for common tangent line of these two circles. The circles touch so there is just one exterior tangent line.

Any help is appreciated, Thanks.

Hello,

you'll find all necessary methods here: http://www.mathhelpforum.com/math-he...712-post1.html
• January 22nd 2008, 01:33 AM
mr fantastic
Quote:

Originally Posted by earboth
Hello,

you'll find all necessary methods here: http://www.mathhelpforum.com/math-he...712-post1.html

Maybe you meant here .....?
• January 22nd 2008, 09:30 AM
Jhevon
Quote:

Originally Posted by mr fantastic
Maybe you meant here .....?

nope. he was almost right the first time, here's the thread. the user double posted
• January 22nd 2008, 09:35 AM
wingless
If you want to get the equation using calculus, follow these steps:

1- Write the circles as an equation. (Like $(x-x_0)^2 + (y-y_0)^2 = r^2$). Then, convert them to functions of x. The thing is, one equation will give you two functions of x, each of them making top and bottom semicircles. Take the functions of top semicircles. For example,
$(x-x_0)^2 + (y-y_0)^2 = r^2$
$(y-y_0)^2 = r^2-(x-x_0)^2$
$|y-y_0| =\sqrt{r^2-(x-x_0)^2}$
$y = y_0 + \sqrt{r^2-(x-x_0)^2}$ (top semicircle), $y = y_0 - \sqrt{r^2-(x-x_0)^2}$ (bottom semicircle)

Now apply it two your circles and take only the functions of top semicircles. Call them f(x), g(x)

2- Let's call the tangent points $(a,f(a))$ and $(b,g(b))$.

3-
i. Slope of the line is $f'(a)$
ii. Slope of the line is $g'(b)$
iii. Slope of the line is $\frac{g(b) - f(a)}{b-a}$

So,
$m = f'(a) = g'(b) = \frac{g(b) - f(a)}{b-a}$

First solve $f'(a) = g'(b)$ and find $b$ as a function of $a$. Then plug this $b$ in $\frac{g(b) - f(a)}{b-a}$ and solve $f'(a) = \frac{g(b) - f(a)}{b-a}$. You'll find $a$. I think you can do the rest easily ^^

This calculations may be hard to do by hand. Using a plotter and a CAS will make your work easier. Good luck ;)

P.S: If you can't solve it, write it here and I or someone else will probably help you wherever you're stuck.
• January 22nd 2008, 11:33 AM
earboth
Quote:

Originally Posted by wingless
If you want to get the equation using calculus, follow these steps:

1- Write the circles as an equation. (Like $(x-x_0)^2 + (y-y_0)^2 = r^2$). Then, convert them to functions of x. The thing is, one equation will give you two functions of x, each of them making top and bottom semicircles. Take the functions of top semicircles. For example,
$(x-x_0)^2 + (y-y_0)^2 = r^2$
$(y-y_0)^2 = r^2-(x-x_0)^2$
$y-y_0 = |r^2-(x-x_0)^2|$
$y = y_0 + r^2-(x-x_0)^2$ (top semicircle), $y = y_0 - r^2-(x-x_0)^2$ (bottom semicircle)

...

Hi,

I'm a little bit confused:

In my opinion this equation $y = y_0 + r^2-(x-x_0)^2$ represents a parabola and not a semi-circle.

Maybe you mean: $y = y_0 + \sqrt{r^2-(x-x_0)^2}$ .....??
• January 22nd 2008, 12:15 PM
wingless
Oh that's a typo, sorry for that, I had to write it and go out quickly :p
It's fixed now.
• January 22nd 2008, 03:26 PM
Henderson
In this situation, the problem is MUCH easier. Since the circles are externally tangent to each other, your tangent line is perpendicular to the segment connecting your centers, and passes through the point $\frac{4}{13}$ of the way across that segment.

Slope of segment:
$\frac{14-2}{9-4} = \frac{12}{5}$

Perpendicular slope:
$-\frac{5}{12}$

Point that the tangent passes through:
$(\frac{4}{13}(2+14),\frac{4}{13}(4+9)) = (\frac{64}{13},4)$

I'll let you find the equation through that point with your perpendicular slope.
• January 23rd 2008, 08:49 PM
CaliMan982
which way?
I am confused which way is the best way to do this cause i got two answers for each, and I do not see where the point 4/13 comes from.
• January 23rd 2008, 10:33 PM
JaneBennet
It seems that Henderson is trying to calculate the internal tangent, that is the one that passes through the point of contact of the two circles. But that’s not the tangent that is wanted, is it? :confused:

In any case, the internal tangent does not pass through $\left(\frac{64}{13},4\right)$. :eek: (It should pass through the point of contact of the circles, which is $\left(2+\frac{4}{13}(14-2),4+\frac{4}{13}(9-4)\right)=\left(\frac{74}{13},\frac{72}{13}\right)$.)