Answer:
Part 12) [tex]Center\ (2,-3),r=2\ units, (x-2)^2+(y+3)^2=4[/tex]
Part 13) [tex]m\angle ABC=47^o[/tex]
Step-by-step explanation:
Part 12) we know that
The equation of a circle in center-radius form is equal to
[tex](x-h)^2+(y-k)^2=r^2[/tex]
where
(h,k) is the center of the circle
r is the radius of the circle
In this problem
Looking at the graph
The center is the [tex]point\ (2,-3)[/tex]
The radius is [tex]r=2\ units[/tex]
substitute in the expression above
[tex](x-2)^2+(y+3)^2=2^2[/tex]
[tex](x-2)^2+(y+3)^2=4[/tex]
Part 13) we know that
The measure of the external angle is the semi-difference of the arcs it covers.
so
[tex]m\angle ABC=\frac{1}{2}[arc\ DE-arc\ AC][/tex]
we have
[tex]arc\ DE=142^o[/tex]
[tex]arc\ AC=48^o[/tex]
[tex]m\angle ABC=\frac{1}{2}[142^o-48^o][/tex]
[tex]m\angle ABC=47^o[/tex]
