Question

30 points I really need your help pronto!

so for the first one, I need help with how to solve the equation. detailed please or not

the second one I understand a bit more I think I have to increase the size or radius by 3 move it 8 units to the right and 2 units up. however, I'm not sure if I should increase the radius by 3 or 4.

the third one I'm just plain lost I don't even know where to start.

Answer 1

Part 1) the center is
`(2,-1)`, the radius is
`4\ units`

Part 2) see the procedure

Part 3)
`m<B=33.2\°`

Explanation:

Part 1) 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

In this problem we have

`(x-2)^(2)+(y+1)^(2) =16`

so

the center is the point
`(2,-1)`

the radius is
`r=√(16)=4\ units`

Part 2) we know that

The center of circle F' is
`(-2,-8)` and the radius is
`r=2\ units`

The center of circle F is
`(6,-6)` and the radius is
`r=4\ units`

step 1

Move the center of the circle F' onto the center of the circle F

the transformation has the following rule

`(x,y)--------> (x+8,y+2)`

8 units right and 2 units up

so

`(-2,-8)--------> (-2+8,-8+2)-----> (6,-6)`

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

`<A+<B+<C+<D=360\°`

substitute the values

`(x+16)+(x)+(6x-4)+(2x+16)=360\°`

solve for x

`(10x+28)=360\°`

`10x=360\°-28\°`

`x=332\°/10=33.2\°`

The measure of angle B is

`m<B=x\°`

so

`m<B=33.2\°`