To prove that Circle [tex]\( X \)[/tex] with radius [tex]\( r \)[/tex] is similar to Circle [tex]\( Y \)[/tex] with radius [tex]\( s \)[/tex], we need to show that there exists a similarity transformation that maps Circle [tex]\( X \)[/tex] to Circle [tex]\( Y \)[/tex]. Here is a detailed step-by-step proof using the given information and logical reasoning:
### Proof:
1. Statement: Circle [tex]\( X \)[/tex] has a radius [tex]\( r \)[/tex].
Reason: Given
2. Statement: Circle [tex]\( Y \)[/tex] has a radius [tex]\( s \)[/tex].
Reason: Given
3. Statement: Translation is a similarity transformation.
Reason: By definition, translation moves every point of a figure or a space by the same distance in a given direction without changing its shape or size.
4. Statement: Dilation is a similarity transformation.
Reason: By definition, dilation multiplies distances from a fixed center (the center of dilation) by a common scale factor, resulting in a shape that is similar to the original.
5. Statement: Translation followed by dilation maps Circle [tex]\( X \)[/tex] to Circle [tex]\( Y \)[/tex].
Reason: A composition of similarity transformations is also a similarity transformation. First, translate Circle [tex]\( X \)[/tex] so that its center coincides with the center of Circle [tex]\( Y \)[/tex]. Next, apply a dilation centered at this common center with a scale factor of [tex]\( \frac{s}{r} \)[/tex].
6. Statement: Circle [tex]\( X \)[/tex] is similar to Circle [tex]\( Y \)[/tex].
Reason: Since we can map Circle [tex]\( X \)[/tex] to Circle [tex]\( Y \)[/tex] using a composition of similarity transformations (translation and dilation), the circles are similar by definition of similarity transformations.
### Conclusion:
Since we have shown that a series of similarity transformations (translation followed by dilation) can map Circle [tex]\( X \)[/tex] to Circle [tex]\( Y \)[/tex], we conclude that Circle [tex]\( X \)[/tex] is similar to Circle [tex]\( Y \)[/tex].
