Answer:
[tex]\sf A. \quad (x, y) \rightarrow \left(\dfrac{7}{2}x, \dfrac{7}{2}y \right)[/tex]
Step-by-step explanation:
The distance from the center of the circle to any point on the circumference is the radius.
From inspection of the given graph:
- Radius of Circle C (before dilation) = 2 units
- Radius of Circle C' (after dilation) = 7 units
To find the scale factor of the dilation from the small circle C to the large circle C', divide the radius of the large circle by the radius of the small circle.
Therefore, the dilation is an enlargement of scale factor ⁷/₂ about the origin.
So the rule that represents the transformation is:
[tex]\sf (x, y) \rightarrow \left(\dfrac{7}{2}x, \dfrac{7}{2}y \right)[/tex]
Check (see attached):
Point (0, 2) is on circle C.
[tex]\implies \sf (0, 2) \rightarrow \left(\dfrac{7}{2}(0), \dfrac{7}{2}(2) \right)=(0,7)[/tex]
Point (1, √3) is on Circle C:
[tex]\implies \sf (1, \sqrt{3}) \rightarrow \left(\dfrac{7}{2}(1), \dfrac{7}{2}(\sqrt{3}) \right)=\left(3.5,6.06)[/tex]
