Question

Circles $P$and $Q$have radii 6 and 2 and are tangent to each other. Find length of their common external tangent $\overline{AB.}$

Short answer

The required length is,

$AB=4\sqrt{5}$.

Step-by-step solution

Step 1. Given information.

Circles $P$and $Q$have radii 6 and 2 and are tangent to each other.

Given figure,

Step 2. Concept used.

There are two circles whose radius is 6 and 2 respectively.

Joint $AP$and $BQ$.

Note that $AP$and $BQ$are radius, then

$\begin{array}{l}AP=6\\ BQ=2\end{array}$

Step 3. Draw the construction.

Joint $PQ$and joint $OB$and make a parallel line to $PQ$. Then$OPQB$ will be a rectangle.

In the rectangle $OPQB$opposite sides are equal.

$\begin{array}{l}OB=PQ\left(=4\right)\\ OP=QB\left(=2\right)\mathrm{.................}\left(1\right)\end{array}$

Then,

$\begin{array}{c}AP=AO+OP\\ 6=AO+2\\ AO=6-2\\ AO=\mathrm{4.............}\left(1\right)\end{array}$

The distance $PQ$is,

$\begin{array}{c}PQ=radius1+radius2\\ =6+2\\ =8\end{array}$

From (1), $OB=PQ$and $PQ=8$

This implies that $OB=8$

Note that $\Delta AOB$is a right-angled triangle.

Step 4. Use Pythagoras theorem.

$\begin{array}{c}A{B}^2=O{A}^2+O{B}^2\\ =4^2+8^2\\ =16+64\\ =80\end{array}$

Take positive square root to get,

$\begin{array}{c}AB=\sqrt{80}\\ =4\sqrt{5}\end{array}$

Therefore, the required length is, $AB=4\sqrt{5}$.