For this case, the equation of the circle is given by:
[tex] (x-1) ^ 2 + (y-3) ^ 2 = 7 ^ 2
[/tex]
We evaluate the points to see which ones satisfy the equation.
We have then:
For (8, 3):
[tex] (8-1) ^ 2 + (3-3) ^ 2 = 7 ^ 2
[/tex]
Rewriting we have:
[tex] (7) ^ 2 + (0) ^ 2 = 7 ^ 2
7 ^ 2 = 7 ^ 2[/tex]
We observe that this point satisfies the equation and therefore, is in the circle.
Answer:
A point lies on the circle represented by the equation [tex](x-1) ^ 2 + (y-3) ^ 2 = 7 ^ 2[/tex] is:
f (8, 3)