check the picture below.
as you know, two tangent lines to the same circle, when they meet, their length is the same, therefore, as you see in the picture, a = a, b = b, c = c, those tangents are three twins.
also recall that the point of tangency, where the radius touches the tangent line, is always a right-angle, as you see there.
and we know the lengths of a + c = 15, a + b = 12 and b + c = 9.
we can write them as a system of equations of 3 variables, and we'll do some elimination, so let's proceed.
![\bf \begin{cases}a+0b+c=15\\a+b+0c=12\\0a+b+c=9\end{cases}\\\\\\\stackrel{\textit{using the first and second equations and multiplying the first one by -1}}{\begin{array}{llcll}a+0b+c=15&* -1\implies &-a-0b-c=-15\\0a+b+c=9&&0a~~+b+c=~~9\\&&---------\\&&-a~~~+b~~~=-6\end{array}}](https://img.qammunity.org/2019/formulas/mathematics/middle-school/xkdmjvow8nxxrl397gfjeki9wknlyrruxl.png)
which we can again write it as -a + b + 0c = -6.
and now, let's use that resultant equation and use it with the second equation as is
![\bf \begin{cases}a+0b+c=15\\a+b+0c=12\\0a+b+c=9\end{cases}\\\\\\\begin{array}{llll}a~~+b+0c=12&\\-a+b+0c=-6\\---------\\~~~~~~2b~~~~~~=6\end{array}\implies b=\cfrac{6}{2}\implies b=3=x](https://img.qammunity.org/2019/formulas/mathematics/middle-school/o96nisuqqlgzshaclg8jfireiqckm7rg1p.png)