we know that
the center of the circle is (4,6) and the radius is 10 units
so
The equation of the circle is given by
[tex]\begin{gathered} (x-4)^2+(y-6)^2=10^2 \\ (x-4)^2+(y-6)^2=100 \end{gathered}[/tex]Remember that
If an ordered pair lie on the circle
then
the ordered pair must satisfy the equation of the circle
Verify each ordered pair
A ---> (-4,12)
For x=-4 and y=12
substitute in the equation of the circle
[tex]\begin{gathered} (-4-4)^2+(12-6)^2=100 \\ (-8)^2+(6)^2=100 \\ 64+36=100 \\ 100=100\text{ ----> is true} \end{gathered}[/tex]point A lies on the circle
B ----> (-6,6)
substitute in the equation
[tex]\begin{gathered} (-6-4)^2+(6-6)^2=100 \\ -10^2+0=100 \\ 100=100 \end{gathered}[/tex]Point B lies on the circle
C -----> (4,6)
Remember that (4,6) is the center
so
Point C does not lie on the circle
D ----> (13,10)
[tex]\begin{gathered} (13-4)^2+(10-6)^2=100 \\ 9^2+4^2=100 \\ 81+16=100\text{ ----> is not true} \end{gathered}[/tex]Point D does not lie on the circle
E -----> (4,16)
[tex]\begin{gathered} (4-4)^2+(16-6)^2=100 \\ 0^2+10^2=100 \\ 100=100 \end{gathered}[/tex]Point F lies on the circle
therefore
The answer are options