# Finding Common Tangent of two circles It seems like easy question, I know all radii are equal in length but I still didn't manage to find BC, any help?

• What have you tried so far? Please edit your question to add in the steps you have taken so far (even if you think they may be pointless) as it will mean people are able to help point out where you may have gone wrong, and then get you going in the right direction – lioness99a Feb 24 at 14:36

Hint:

Consider the height and the hypotenuse of the right triangle. • Thanks for the information, I now managed to solve it thanks! – Abdallah Hamad Feb 24 at 14:45

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}

• Pack.Nice.Hypothenuse law = Pythagoras' theorem. – Peter Szilas Feb 24 at 14:48

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 $$T^2+(r_1-r_2)^2=(r_1+r_2)^2$$

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

• This isn't very helpful as it currently stands - please add some explanation to help the OP understand where you got this formula from – lioness99a Feb 24 at 14:37