# SAT Mathematics : Applying the Equation of a Circle

## Example Questions

### Example Question #1 : Applying The Equation Of A Circle

What is the equation for a circle of radius 12, centered at the intersection of the two lines:

and

?

Explanation:

To begin, let us determine the point of intersection of these two lines by setting the equations equal to each other:

To find the y-coordinate, substitute into one of the equations. Let's use :

The center of our circle is therefore: (–41, –161).

Now, recall that the general form for a circle with center at  is:

For our data, this means that our equation is:

or

### Example Question #1 : Applying The Equation Of A Circle

What is the radius of a circle with the equation ?

Explanation:

We need to expand this equation to  and then complete the square.

This brings us to .

We simplify this to .

### Example Question #5 : Applying The Equation Of A Circle

A circle has its origin at . The point  is on the edge of the circle. What is the radius of the circle?

There is not enough information to answer this question.

Explanation:

The radius of the circle is equal to the hypotenuse of a right triangle with sides of lengths 5 and 7.

This radical cannot be reduced further.

### Example Question #2 : Applying The Equation Of A Circle

A circle with a radius of five is centered at the origin. A point on the circumference of the circle has an x-coordinate of two and a positive y-coordinate. What is the value of the y-coordinate?

Explanation:

Recall that the general form of the equation of a circle centered at the origin is:

We know that the radius of our circle is five. Therefore, we know that the equation for our circle is:

Now, the question asks for the positive y-coordinate when .  To solve this, simply plugin for :

Since our answer will be positive, it must be .

### Example Question #1 : Applying The Equation Of A Circle

The following circle is moved  spaces to the left. Where is its new center located?

Explanation:

Remember that the general equation for a circle with center  and radius  is

With that in mind, our original center is at  .

If we move the center  units to the left, that means that we are subtracting  from our given coordinates.

Therefore, our new center is .

### Example Question #1 : Applying The Equation Of A Circle

A square on the coordinate plane has vertices at the points with coordinates , and . Give the equation of the circle that circumscribes the square.

Explanation:

The equation of the circle on the coordinate plane with radius  and center  is

The figure referenced is below:

The center of the circle is at the point of intersection of the diagonals, which, as is the case with any rectangle, bisect each other. Therefore,  looking at the diagonal with endpoints  and , we can set  in the midpoint formula:

and

The center of the circumscribing circle is therefore .

The radius of the circumscribing circle is the distance from this point to any point on the circle. The distance formula can be used here:

Since we are actually trying to find , we will use the form

Choosing the radius with endpoints  and , we set  and substitute:

Setting  and  and substituting in the circle equation:

, the correct response.

### Example Question #2 : Applying The Equation Of A Circle

The above figure shows a circle on the coordinate axes with its center at the origin.  has length

Give the equation of the circle.

Explanation:

arc of a circle represents  of the circle, so the length of the arc is three-eighths its circumference. Set up the equation and solve for

The equation of a circle on the coordinate plane is

,

where  are the coordinates of the center and  is the radius.

The radius of a circle can be determined by dividing its circumference by , so

The center of the circle is , so . Substituting 0, 0, and 8  for , and , respectively, the equation of the circle becomes

,

or

.

### Example Question #10 : Applying The Equation Of A Circle

The above figure shows a circle on the coordinate axes with its center at the origin. The shaded region has area .

Give the equation of the circle.

Explanation:

The unshaded region is a  sector of the circle, making the shaded region a  sector, which represents  of the circle. Therefore, if  is the area of the circle, the area of the sector is . The sector has area , so

Solve for :

The equation of a circle on the coordinate plane is

,

where  are the coordinates of the center and  is the radius.

The formula for the area  of a circle, given its radius , is

.

Set  and solve for :

The center of the circle is , so . Substituting 0, 0, and 56 for , and , respectively, the equation of the circle becomes

,

or

.

### Example Question #2 : Applying The Equation Of A Circle

If the center of a circle is at (0,4) and the diameter of the circle is 6, what is the equation of that circle?

Explanation:

The formula for the equation of a circle is:

Where (h,k) is the center of the circle.

and

and diameter = 6 therefore radius = 3

### Example Question #5 : Applying The Equation Of A Circle

Circle A is given by the equation . Circle A is shifted up five units and left by six units. Then, its radius is doubled. What is the new equation for circle A?

Explanation:

The general equation of a circle is , where (h, k) represents the location of the circle's center, and r represents the length of its radius.

Circle A first has the equation of . This means that its center must be located at (4, –3), and its radius is .

We are then told that circle A is shifted up five units and then left by six units. This means that the y-coordinate of the center would increase by five, and the x-coordinate of the center would decrease by 6. Thus, the new center would be located at (4 – 6, –3 + 5), or (–2, 2).

We are then told that the radius of circle A is doubled, which means its new radius is .

Now, that we have circle A's new center and radius, we can write its general equation using .

.

.