Question

prove that all circles are similar using two circles, \\( \odot a \\) with a radius of \\( r=a \\) and \\( \odot b \\) with a radius of \\( r = b \\).
move the answers to complete the sentences.
using the circumference formula \\( c = 2\pi r \\), the circumference of \\( \odot a \\) is \\( c_a = 2\pi a \\) and the circumference of \\( \odot b \\) is \\( c_b = 2\pi b \\).
then use the division property of equality for each equation to rewrite the equations with a ratio of the circumference to the radius. for \\( \odot a \\) this gives us and for \\( \odot b \\) this gives us
now use the transitive property of equality to get the equation
we can rewrite the proportion to have the ratio of the radius of \\( \odot a \\) to the radius of

Explanation:

Step1: Find ratio for circle A

Given \(C_A = 2\pi a\), divide both sides by \(a\).
\(\frac{C_A}{a}=\frac{2\pi a}{a}=2\pi\)

Step2: Find ratio for circle B

Given \(C_B = 2\pi b\), divide both sides by \(b\).
\(\frac{C_B}{b}=\frac{2\pi b}{b}=2\pi\)

Step3: Use transitive property

Since \(\frac{C_A}{a}=2\pi\) and \(\frac{C_B}{b}=2\pi\), by transitive property \(\frac{C_A}{a}=\frac{C_B}{b}\)

Answer:

For \(\odot A\): \(\frac{C_A}{a} = 2\pi\); for \(\odot B\): \(\frac{C_B}{b}=2\pi\); transitive property equation: \(\frac{C_A}{a}=\frac{C_B}{b}\)