Basic Pi related question - comparing circles

**ZenoZammo** · November 26th 2011, 04:07 PM

Hi,

Just joined this forum as trying to get back into mathematics after not doing much in this subject since school. I'm going back over the basics to reacquaint myself with the tools I'll need to progress, and thought I'd play around with Pi for some fun. Whilst doing so, found a pattern that I'm sure will be obvious to many of you (probably just some easy algebraic manipulation I'm overlooking), so thought I'd post here to ask for your help explaining.

Was looking into radius, area and circumference of circles. Thought about what if you had two circles, one with radius r, and the second with a circumference of r-squared (for example, first circle had radius 5, second circle had circumference 25).

Found patterns in the ratios between the circles, when comparing radius, area and circumference. See results below:

Circle 1:
...Radius: 5
...Area: 78.53982
...Circumference: 31.41592654

Circle 2:
...Radius: 3.978874
...Area: 49.73592
...Circumference: 25

Ratio between circle metrics (Circle 2/Circle 1):
...Radius: 1.256637
...Area: 1.579137 (which is approximately 1.256637 squared)
...Circumference: 1.256637061

What interests me is why the ratios are so linked, I'm sure there's a simple algebraic explanation, but due to my rustiness I'm unsure where to start. Can anyone here help me?

Thanks in advance!

**emakarov** · November 26th 2011, 04:33 PM

Welcome to the forum.

Let $r_i$ , $c_i$ and $a_i$ denote radius, circumference and area of circle $i$ ( $i=1,2$ ). Then $c_i=2\pi r_i$ and $a_i=\pi r_i^2$ . Also, $c_2=r_1^2$ by assumption, so $r_2=\frac{r_1^2}{2\pi}$ and $a_2=\pi r_2^2=\frac{r_1^4}{4\pi}$ .

Therefore, $\frac{r_2}{r_1}=\frac{r_1}{2\pi}$ and $\frac{a_2}{a_1}=\frac{r_1^4}{4\pi} \cdot \frac{1}{\pi r_1^2}=\frac{r_1^2}{4\pi^2}$ .

**awkward** · November 26th 2011, 04:38 PM

Hi ZenoZammo,

Let r = the radius of a circle, C = the circumference, and A = the area. Then we know the classic formulas
$C = 2 \pi r$
and
$A = \pi r^2$

So if we have two circles of radii $r_1$ and $r_2$ , then the circumference and area of the first circle are
$C_1 = 2 \pi r_1$
and
$A_1 = \pi r_1^2$
and the circumference and area of the second circle are
$C_2 = 2 \pi r_2$
and
$A_2 = \pi r_2^2$

So
$\frac{C_2}{C_1} = \frac{2 \pi r_2}{2 \pi r_1} = \frac{r_2}{r_1}$
and
$\frac{A_2}{A_1} = \frac{\pi r_2^2}{\pi r_1^2} = \left( \frac{r_2}{r_1} \right) ^2$

**ZenoZammo** · November 27th 2011, 02:15 AM

Thank you to you both, emakarov and awkward, I appreciate you taking the time to answer my question.

awkward, I found your explanation easier to grasp. emakarov, I found your explanation useful too as it reminded me of some algebraic manipulation techniques I had forgotten.

Off to find my next maths challenge, thanks again. :-)

**emakarov** · November 27th 2011, 07:34 AM

You are welcome. You are right that the explanation in post #3 is easier. I did not realize right away that $c_2/c_1=r_2/r_1$ and $a_2/a_1=(r_2/r_1)^2$ for any circle. The fact that $c_2=r_1^2$ is not used here.

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