Why would we use the radius of a circle instead of the diameter when calculating circumference? Forgive me if this question is a little too strange or maybe even off.  Mathematics has never been my strong point, but I definitely think it's the coolest...
Anyway, I was looking into tau, pi's up-and-coming sibling. I started rethinking why pi worked. The thing about tau is that it supposedly skips the step of doubling πr, since tau is twice pi. I tried this, and it works! (Of course it works. The fact that this surprises and amazes me shows how little I get out...)
Then I started wondering why we do 2πr instead of πd. It does give the same answer... I checked. Is there any reason why using the radius is preferred over using the diameter?
Here's my work:

r = 6 (ergo diameter must equal 12)
C = 2πr
C = 37.68
and...
C = Tr
C = 37.68
and...
C = πd
C = 37.68


You can define a circle knowing the centre and the radius (distance $r$).   A circle is the set of all points, on a 2D-plane, at distance $r$ from the centre.
That's a concise and elegant definition; try doing so using the diameter (distance $d$).
Then, having defined circles using the radius, it becomes convenient to also define the radian measure of angles in terms of the radius of a circle.   One radian is the measure of an angle subtended at the centre of a circle by an arc length equal to the radius.
Then we asked: what is the radian measure of a straight angle (that formed by two rays of a line)?   Why it is that irrational number we have decided to call $\pi$, to honour Pythagorus.
What then is the angle subtended by the circumference of a circle?   Well, we could call it $\tau$, but it is $2\pi$, and we just happened to have named $\pi$ first.
Thus $$\begin{array}{cc}C&=&2\pi r &=& \pi d &=& \tfrac 12 \tau d &=& \tau r & \text{circumference of circle}\\[1ex]A&=& \pi r^2 &=& \tfrac 14 \pi d^2 &=& \tfrac 18 \tau d^2 &=& \tfrac 12\tau r^2 & \text{area of circle/disc} \\[2ex] S &=& 4\pi r^2 &=& \pi d^2 &=&\tfrac 12 \tau d^2 &=&2\tau r^2 & \text{surface area of sphere}\\[1ex] V &=& \tfrac 43 \pi r^3 &=& \tfrac 16 \pi d^3 &=&\tfrac 1{12}\tau d^3 &=& \tfrac 23 \tau r^3 & \text{volume of sphere/ball}\end{array}$$
