# Common external tangent

• Jan 18th 2010, 08:57 PM
reiward
Common external tangent

Two circles have the radii 6 and 3 cm. The distance between their centers is 18cm. Find the length of the common external tangent.

See figure.
• Jan 19th 2010, 01:45 AM
Hello reiward
Quote:

Originally Posted by reiward

Two circles have the radii 6 and 3 cm. The distance between their centers is 18cm. Find the length of the common external tangent.

See figure.

I think the diagram should look like the one I have attached.

So, in this diagram, the angles at $R$, $S$ and $T$ are all right angles. $QR = TS = 3$ cm; $PS = 6$ cm; $PQ = 18$ cm. We need to calculate $SR$, which is the same length as $TQ$.

Can you complete it now?

Hint:
Spoiler:
Use Pythagoras on $\triangle PQT: PQ^2= PT^2+TQ^2$.

• Jan 19th 2010, 02:03 AM
reiward
Uhmm, the diagram I attached was provided by my teacher, yours is I think different.
• Jan 19th 2010, 03:19 AM
Hello reiward
Quote:

Originally Posted by reiward
Uhmm, the diagram I attached was provided by my teacher, yours is I think different.

I think you'll find that's the common internal tangent.

But if you need its length, draw the line joining the centres of the circles. Then use similar triangles, with one pair of corresponding sides being radii of the two circles. So the tangent crosses the line joining the centres at a point dividing it in the ratio 2:1; the distances from the centres of this point therefore being 12 cm and 6 cm.

Then use Pythagoras on the triangles to find the lengths of the two parts of the tangent. Add together. Answer: 15.59 cm

• Jan 19th 2010, 12:01 PM
earboth
Quote:

Originally Posted by reiward

Two circles have the radii 6 and 3 cm. The distance between their centers is 18cm. Find the length of the common external tangent.

See figure.

Here comes a slightly different attempt:

1. Draw a circle around $M_2$ with radius $R = r_1+r_2$.

2. Draw the tangent from $M_1$ to the new, greater circle. (Red line).

3. This red line segment is parallel to the internal tangent and has the same length as the internal tangent.

4. The indicated triangle is a right triangle with hypotenuse $|\overline{M_1 M_2}|$ and one leg $|\overline{r_1+r_2}|$. Use Pythagorean theorem to calculate the length of the internal tangent..

$l_t = \sqrt{18^2-9^2} = 9\sqrt{3}$