According to Wikipedia, pi is "a mathematical constant that was originally defined as the ratio of a circle's circumference to its diameter."
However, when I think of the word "ratio", something like $4:3$ or $7:10$ comes to mind. Why is pi said to be a ratio? Isn't it more accurate to say that pi is the circumference divided by its diameter, rather than the ratio of the circumference to the diameter (which I would think of as $\pi:1$)?
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Recall that circumference of a circle is $C = \pi d$, where $d$ is the diameter. Then, $$\pi={C\over d},$$ as you mentioned. If we had the exact length of the circumference of the circle, and its diameter, then we will compute $\pi$ exactly. However, as you learn in any science class, our instruments of measurement are not perfect and subject to roundoff error. So maybe using a ruler, you compute that your circle's circumference is $31$ inches, and its diameter is $10$ inches. Then $\pi \approx 31 \div10=3.1$. If we can get a more precise measuring tool, we get closer to $\pi=3.141592\ldots$
The phrase "ratio of $a$ and $b$" just means "$a$ divided by $b$." They are synonymous. I understand you said that you think of "ratio" and "division" differently, but that is a mistake. Both the symbols $a:b$ and $a/b$ mean the same thing. Although, to be honest with you, once you enter higher level education, you will literally never encounter the former notation. The only notation that is used is $a/b$ or $\frac{a}{b}.$ "Ratio" is just another word for talking about division.
A better way to understand the definition being given is that, it can be proven that for all circles in a Euclidean space, the circumference $C$ is proportional to the diameter $d,$ which is to say, there exists some constant quantity $\pi$ such that $C=\pi\cdot{d}$ for all circles in Euclidean space, and this quantity $\pi$ is independent of $d,$ because it is constant and fixed. This quantity $\pi$ can be understood to be a real number. If you use calculus to compute the circumference of an arbitrary circle, you can eventually prove that $$C=d\int_{-1}^1\frac1{\sqrt{1-x^2}}\,\mathrm{d}x,$$ and $$\int_{-1}^1\frac1{\sqrt{1-x^2}}\,\mathrm{d}x$$ is indeed constant. So we say $$\pi=\int_{-1}^1\frac1{\sqrt{1-x^2}}\,\mathrm{d}x.$$ However, this is not the definition we actually use in higher level mathematics.