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Question
- lin is trying to convince andre that all circles are similar. help her write a valid justification for why all circles are similar.
To show all circles are similar, recall the definition of similar figures: figures that can be mapped onto each other via a similarity transformation (translation, rotation, reflection, dilation). For circles, any circle can be transformed into another using dilation. Let’s take two circles with radii \( r_1 \) and \( r_2 \). The ratio of their radii is \( \frac{r_2}{r_1} \) (or \( \frac{r_1}{r_2} \)). A dilation with scale factor \( k=\frac{r_2}{r_1} \) (centered at any point) will map the first circle (radius \( r_1 \)) to the second circle (radius \( r_2 \)). Since dilation is a similarity transformation, and all circles have the same shape (defined by the constant \( \pi \) in their circumference/area formulas, and their set of points equidistant from a center), this shows all circles are similar. Also, the angle measures in a circle (e.g., central angles, inscribed angles) are consistent in their proportionality when scaled, and the only dimension defining a circle is its radius (or diameter), so scaling the radius scales the entire circle uniformly without changing its shape.
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All circles are similar because any circle can be transformed into another via a dilation (a similarity transformation). For two circles with radii \( r_1 \) and \( r_2 \), a dilation with scale factor \( \boldsymbol{\frac{r_2}{r_1}} \) (or \( \frac{r_1}{r_2} \)) maps one circle to the other. Circles have the same shape (defined by the constant relationship between radius, circumference, and area, e.g., \( C = 2\pi r \), \( A=\pi r^2 \)), so scaling the radius (via dilation) preserves their similarity.