Question

Need help with two questions I am not good with this

Answer 1

Answer:

Part 12) $Center\ (2,-3),r=2\ units, (x-2)^2+(y+3)^2=4$

Part 13) $m\angle ABC=47^o$

Step-by-step explanation:

Part 12) we know that

The equation of a circle in center-radius form is equal to

$(x-h)^2+(y-k)^2=r^2$

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 $point\ (2,-3)$

The radius is $r=2\ units$

substitute in the expression above

$(x-2)^2+(y+3)^2=2^2$

$(x-2)^2+(y+3)^2=4$

Part 13) we know that

The measure of the external angle is the semi-difference of the arcs it covers.

so

$m\angle ABC=\frac{1}{2}[arc\ DE-arc\ AC]$

we have

$arc\ DE=142^o$

$arc\ AC=48^o$

$m\angle ABC=\frac{1}{2}[142^o-48^o]$

$m\angle ABC=47^o$