The point is the center of the circle - it is not on the circle.

We can test to see if the other points are actually on the circle by plugging in their x and y values into the equation. For example, to verify that

is actually on the circle, we can plug in  for x and for y:

this is true, so that point is on the circle.

To quickly figure this out, we can plug in 5 for x and -2 for y and see what happens:

Since the value is greater than 9, this point is outside the radius of this circle.

Start with the equation

It would be helpful if we "complete the square" for both the x and y polynomials.

So by adding 9 and 900 we transform the equation into

And then by dividing 900, we see that the outcome is:

The standard equation for an ellipse is

With a center at 

Therefore this is the equation of an ellipse, with center 

Remember that the equation for an ellipse in standard form looks like the following:

Where the point (h,k) gives the center of the ellipse, a is half the length of its axis in the x direction, and b is half the length of its axis in the y direction. We can see that this form has a 1 on the right side of the equation, so let's start by dividing both sides of our equation by 36 to get a 1 on the right side:

Now we can simplify the fractions on the right side of the equation, which gives us the equation for our ellipse in standard form:

This ellipse would have its center at (4,7), would be 6 units wide in the x direction, and 4 units wide in the y direction, because so , and so .

Remember that in order for the equation of an ellipse to be written in standard form, it must be written in one of the following two ways:

Where the point (h,k) gives the center of the ellipse, a is half the length of the axis for which it is the denominator, and b is half the length of the axis for which it is the denominator. Looking at our answer choices, we need one that involves addition, has different denominators under the x and y terms, and is equal to 1 on the right side of the equation. We can see that out of all the answer choices, the following is the only one that satisfies those requirements of standard form:

The equation of an ellipse in standard form is

,

where  is the center point,  is half the length of its axis in the  direction, and  is half the length of the axis in the  direction.

In this instance, the center point  , , and . Therefore:

Recall that the general equation of a circle is: 

. 

The key feature here being that both the  and  terms are squared.

Thus, we can narrow down the choices without this feature.

Next, consider the center. Since the center here is , we should not have any middle terms.

The only choice that fits that description is . 

The general equation for an ellipse is , although if we consider a to be half the length of the major axis, a and b might switch depending on if the longer major axis is horizontal or vertical. This general equation has as the center, a as the length of half the major axis, and b as the length of half the minor axis.

Because the center of this ellipse is at and the foci are at , we can see that the foci are away from the center, and they are on the horizontal axis. This means that the horizontal axis is the major axis, the one with length 14. Having a length of 14 means that half is 7, so . Since the foci are away from the center, we know that . We can solve for b using the equation :

that's really as far as we need to solve.

Putting all this information into the equation gives:

The formula for an ellipse centered at the point  with a horizontal major axis (ie: parallel to the x-axis) has the formula

where  and  are the lengths of the major and minor axes respectively.

Since the origin is at  and  and  in this problem, the equation is

or

Because our equation is already in the format

 ,

we do not have to manipulate the equation. The center of any ellipse in this form will always be . So in this case, our center will be . To find the foci of the ellipse, we must use the equation , where  is the greater of the two denominators in our equation ( and ),  is the lesser and  is the distance from the center to the foci.

We know that  and .

By using , we see that , so .

We now know that the two foci are going to be  units in either direction of the center along the greater axis.

Because the greater denominator is under our term containing , our ellipse will have its greater axis going vertically, rather than horizontally. Therefore, our foci will be  units above and below our center, at  and .