Finding Common Tangent of two circles

Question

It seems like easy question, I know all radii are equal in length but I still didn't manage to find BC, any help?

Answer 1

Hint:

Consider the height and the hypotenuse of the right triangle.

Answer 2

When two circlesof radii $r_1$ and $r_2$ touch each other externally then by pythagorous theorem the length of direc common tangent is given by $T^2+(r_1-r_2)^2=(r_1+r_2)^2$

$$BC=T=\sqrt{(8+5)^2-(8-5)^2}= 2 \sqrt{8\times 5}=4\sqrt{10}$$

Answer 3

Consider the triangle $MNO$ with $O$ being the orthogonal projection of $N$ on the segment $[MC]$.

The length of $MC$ is $8$ centimeters. The length of $CO$ is $5$ centimeters.

Therefore the length of $OM$ is $3$ centimeters. Also, you know the length $$MN = MA + AN = 8 + 5$$

Now with the hypotenuse law, you can find that \begin{align}MN^2 &= MO^2 + ON^2 \\ 3^2 + x^2 &= (8+5)^2\end{align}