# Circles and Internal Tangent Segments

• June 17th 2011, 04:33 PM
Zellator
Circles and Internal Tangent Segments
Hi Forum.
I have been studying secant and tangent theorems for the circles and I found this question:

Find the length of the common internal tangent segment of two circles with radii of 4 units and 12 units whose centers are 20 units apart.

Ok, I drew a sketch and I tried to used secant tangent theorems but I didn't get it off the ground.
There's a formula that states $x=\sqrt{y^2-(r_1+r_2)^2}$
With $y$ being the segment's length.
I can only think of ways of solving this, cutting the segment into two parts, that are not equal, and then summing them.
But I don't know how to put the distance of 20 units between the centers.

My question is, how do we get the formula $x=\sqrt{y^2-(r_1+r_2)^2}$?

Thanks!

Note that the x and the y in the formula are not the same x and y in the sketch.
Here is a sketch:
• June 17th 2011, 06:04 PM
Also sprach Zarathustra
Re: Circles and Internal Tangent Segments
Quote:

Originally Posted by Zellator
Hi Forum.
I have been studying secant and tangent theorems for the circles and I found this question:

Find the length of the common internal tangent segment of two circles with radii of 4 units and 12 units whose centers are 20 units apart.

Ok, I drew a sketch and I tried to used secant tangent theorems but I didn't get it off the ground.
There's a formula that states $x=\sqrt{y^2-(r_1+r_2)^2}$
With $y$ being the segment's length.
I can only think of ways of solving this, cutting the segment into two parts, that are not equal, and then summing them.
But I don't know how to put the distance of 20 units between the centers.

My question is, how do we get the formula $x=\sqrt{y^2-(r_1+r_2)^2}$?

Thanks!

Note that the x and the y in the formula are not the same x and y in the sketch.
Here is a sketch:

From Pythagoras theorem:

(1) $12^2+t_2^2 =(x+12)^2$

(2) $4^2+t_1^2=(y+4)^2$

Now, you now that (3) $x+y=4$ (why?)

And also you have two similar triangles, so:

(4) $\frac{t_1}{t_2}=\frac{4}{12}$

All the above (1), (2), (3) and (4) should be enough for you solving the problem...

EDIT:

You can arrive to your formula by 'playing' with the above equations...(try it by your own)
• June 18th 2011, 12:56 AM
earboth
Re: Circles and Internal Tangent Segments
Quote:

Originally Posted by Zellator
Hi Forum.
I have been studying secant and tangent theorems for the circles and I found this question:

Find the length of the common internal tangent segment of two circles with radii of 4 units and 12 units whose centers are 20 units apart.

...

1. I've modified your sketch a little bit (see attachment).

2. Use proportions with the similar (greyed) triangles:

$\dfrac{t_1}{t_2}=\dfrac{r_1}{r_2}~\implies~t_1 = \frac{r_1}{r_2} \cdot t_2$

3. Since you know the distance between the centers and the lengthes of the radii you can use Pythagorean theorem to determine $t_1+t_2 = 12$.

4. Replace $t_1$ in the last equation by the term of #2.
Solve for $t_2$ and afterwards for $t_1$.
• June 18th 2011, 02:21 AM
bjhopper
Re: Circles and Internal Tangent Segments
Hi Zellator,
You can solve this problem using your sketch.

x+y = 4
4/12 =y+4/x+12
whence x= 3 y=1
(12+x)^2 =4^2 +(t2)^2
15^2 = 4^2 + (t2)^2
t2 =9
t1/t2 = 1/3
t1 =3

bjh
• June 18th 2011, 12:33 PM
Zellator
Re: Circles and Internal Tangent Segments
Hey Guys!