Question

A circle is translated 4 units to the right and then reflected over the x-axis. Complete the statement so that it will always be true of the circle at the new location.

Answer 1

the same area as

Step-by-step explanation:

When a circle is translated and reflected, the center of the circle will change; however, its area, circumference, radius and diameter remain the same.

This is so because, translation and reflection only affect the positioning of the circle not the size.

Considering the above analysis, we can conclude that option d answers the question correctly.

Answer 2

The statement is now presented as:

`\exists\, (h,k)\in \mathbb{R}^(2) /f: (x-h^(2))+(y-k)^(2)=r^(2)\implies f': [x-(h+4)]^(2)+[y-(-k)]^(2) = r^(2)`

In other words, this mathematical statement can be translated as:

There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r.

Explanation:

Let
`C = (h,k)` the coordinates of the center of the circle, which must be transformed into
`C'=(h', k')` by operations of translation and reflection. From Analytical Geometry we understand that circles are represented by the following equation:

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

Where
`r` is the radius of the circle, which remains unchanged in every operation.

Now we proceed to describe the series of operations:

1) Center of the circle is translated 4 units to the right (+x direction):

`C''(x,y) = C(x, y) + U(x,y)` (Eq. 1)

Where
`U(x,y)` is the translation vector, dimensionless.

If we know that
`C(x, y) = (h,k)` and
`U(x,y) = (4, 0)`, then:

`C''(x,y) = (h,k)+(4,0)`

`C''(x,y) =(h+4,k)`

2) Reflection over the x-axis:

`C'(x,y) = O(x,y) - [C''(x,y)-O(x,y)]` (Eq. 2)

Where
`O(x,y)` is the reflection point, dimensionless.

If we know that
`O(x,y) = (h+4,0)` and
`C''(x,y) =(h+4,k)`, the new point is:

`C'(x,y) = (h+4,0)-[(h+4,k)-(h+4,0)]`

`C'(x,y) = (h+4, 0)-(0,k)`

`C'(x,y) = (h+4, -k)`

And thus,
`h' = h+4` and
`k' = -k`. The statement is now presented as:

`\exists\, (h,k)\in \mathbb{R}^(2) /f: (x-h^(2))+(y-k)^(2)=r^(2)\implies f': [x-(h+4)]^(2)+[y-(-k)]^(2) = r^(2)`

In other words, this mathematical statement can be translated as:

There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r.