The image of a point  under a dilation with center  and scale factor  is the point in the same direction from  as , but  units away, where  is the distance from  to . Since the center of the dilation is the origin , the coordinates of the image of , which is , can be found by multiplying scale factor  by each of those of :

The images of the other three vertices can be found similarly:

Note that only in the case of , the ordered pair in the choice differs (in abscissa) from the correct ordered pair. This makes the correct choice .

To perform a dilation with center  and scale factor , find the midpoints of the segments connecting  to each point, and connect those points. We can simplify the process by finding the midpoints of , , and , and naming them , , and , respectively; , the image of center , is just  itself. The figure is below:

Now, do the same thing to the new square, but with  as the center. The figure is below:

The final image, relative to the original square, is below:

The centroid of a triangle can be located by finding the intersection of the three medians of the triangle - the segments that connect each vertex to the midpoint of its opposite side. The medians are shown below, with point of intersection  :

A dilation of scale factor   with center  can be performed by letting , , and  be the midpoints of , , and , respectively: 

Removing the medians and , we see that the correct choice is the figure

The graph of the equation

is an ellipse whose center is at the origin; it has a horizontal axis with endpoints at  and a vertical axis with endpoints at  . Thus,

,

or

has its center at the origin, and its endpoints at  and .

The center of dilation is the center of the ellipse, so the center of the image will also be at the origin. The endpoints of the ellipse will be the points  times the distance away from the origin, in the same directions. Since the center of the dilation is the origin, these points can be found by simply multiplying the coordinates by ; the endpoints of the horizontal axis will be

,

or

,

and the endpoints of the vertical axis will be

or

.

This dilation is shown in the figure below:

Therefore,  and , making the equation of the image

,

or

.

The orthocenter of a triangle can be located by finding the intersection of the three altitudes of the triangle - the segments connecting each vertex to its opposite side, perpendicular to the respective side. Since the triangle is right,  and  are two of the altitudes, which intersect at ; the third altitude must also pass through , since the three altitudes are concurrent. Therefore, we perform a dilation of the triangle with respect to center .

This is done by mapping  and  to the midpoints of  and , respectively, and by mapping  to itself. The triangle is seen below:

This figure is the correct choice.

To perform a dilation with center  and scale factor , find the midpoints of the segments connecting  to each point, and connect those points. We can simplify the process by finding the midpoints of , , and , and naming them , , and , respectively; , the image of center , is just  itself. The figure is below:

Now, do the same thing to the new square, but with  as the center. The figure is below:

The image, relative to the original square, is below:

The graph of the equation

is an ellipse whose center is at the origin; it has a horizontal axis with endpoints at  and a vertical axis with endpoints at  . Thus,

,

or

,

has its center at the origin, and its endpoints at and .

The center of dilation is the center of the ellipse, so the center of the image will also be at the origin. The endpoints of the ellipse will be the points 3 times the distance away from the origin, in the same directions. Since the center of the dilation is the origin, these points can be found by simply multiplying the coordinates by 3; the endpoints of the horizontal axis will be

,

or

,

and the endpoints of the vertical axis will be

or

.

The figure is shown below:

Therefore, and , making the equation of the image

,

or

.

The orthocenter of a triangle can be located by finding the intersection of the three altitudes of the triangle - the segments connecting each vertex to the line containing its opposite side, perpendicular to the respective side. Extend  and  to  and ;  the altitudes are shown below, with point of intersection  :

A dilation of scale factor   with center  can be performed by letting , , and  be the midpoints of , , and , respectively: 

Removing the altitudes and , we see that the correct choice is the figure:

The graph of the equation

is a circle with center  and radius .

,

or

is a circle with  center at origin  and radius 8.

A dilation of a circle with scale factor  will result in multiplying that radius by , so the radius of the circle will be . 

To find the center of the image, note that the origin is 4 units below . The center of the new circle must be  units below , so this center will be , or . See the figure below:

Substituting in the circle formula, this is 

,

or

.

The circumcenter of a triangle can be located by finding the intersection of the perpendicular bisectors of the three sides of the triangle. The perpendicular bisectors are shown below, with point of intersection  :

Construct , , and . A dilation of scale factor   with center  can be performed by letting , , and  be the midpoints of , , and , respectively: 

Removing the perpendicular bisectors and , we see that the correct choice is the figure