Answer:
It can be proved that the circle R is similar to the circle Q by translating the circle R a displacement of (-6, 12).
Step-by-step explanation:
We can demonstrate that Circle R is similar to Circle Q by translating the center of the former one to the center of latter one. Meaning that every point of circle R experiments the same translation. Vectorially speaking, a translation is defined by:
[tex]O'(x,y) = O(x,y) + T(x,y)[/tex] (1)
Where:
[tex]O(x,y)[/tex] - Original point.
[tex]O'(x,y)[/tex] - Translated point.
[tex]T(x,y)[/tex] - Translation vector.
If we know that [tex]O(x,y) = (2,-9)[/tex] and [tex]O'(x,y) = (-4,3)[/tex], then the translation vector is:
[tex]T(x,y) = O'(x,y)-O(x,y)[/tex]
[tex]T(x,y) = (-4,3) - (2,-9)[/tex]
[tex]T(x,y) = (-6,12)[/tex]
It can be proved that the circle R is similar to the circle Q by translating the circle R a displacement of (-6, 12).
