Answer:
Part 1) the center is [tex](2,-1)[/tex], the radius is [tex]4\ units[/tex]
Part 2) see the procedure
Part 3) [tex]m<B=33.2\°[/tex]
Step-by-step explanation:
Part 1) 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
In this problem we have
[tex](x-2)^{2}+(y+1)^{2} =16[/tex]
so
the center is the point [tex](2,-1)[/tex]
the radius is [tex]r=\sqrt{16}=4\ units[/tex]
Part 2) we know that
The center of circle F' is[tex](-2,-8)[/tex] and the radius is [tex]r=2\ units[/tex]
The center of circle F is[tex](6,-6)[/tex] and the radius is [tex]r=4\ units[/tex]
step 1
Move the center of the circle F' onto the center of the circle F
the transformation has the following rule
[tex](x,y)--------> (x+8,y+2)[/tex]
8 units right and 2 units up
so
[tex](-2,-8)--------> (-2+8,-8+2)-----> (6,-6)[/tex]
center circle F' is now equal to center circle F
The circles are now concentric (they have the same center)
step 2
A dilation is needed to increase the size of circle F' to coincide with circle F
scale factor=radius circle F/radius circle F'=4/2=2
radius circle F' will be=2*scale factor=2*2=4 units
radius circle F' is now equal to radius circle F
A translation, followed by a dilation will map one circle onto the other
Part 3) we know that
The sum of the interior angles in a quadrilateral is equal to 360 degrees
so
[tex]<A+<B+<C+<D=360\°[/tex]
substitute the values
[tex](x+16)+(x)+(6x-4)+(2x+16)=360\°[/tex]
solve for x
[tex](10x+28)=360\°[/tex]
[tex]10x=360\°-28\°[/tex]
[tex]x=332\°/10=33.2\°[/tex]
The measure of angle B is
[tex]m<B=x\°[/tex]
so
[tex]m<B=33.2\°[/tex]
