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

.

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

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 .

.

.

The answer is .

The diameter of Circle B will be the diagonal of the small Square A, but will be the length of the large Square C. You will use the diagonal information to determine the side of the small Square A, then you can calculate the area of each.

First, let's find the side of small Square A. If the radius of Circle B is , then the diameter is . You can use the diagonal to view Square A as two right isosceles () triangles. Since the hypotenuse of the triangle is , then the side of the triangle, and thus the side of Square A, is . The area of this Square A is , which simplifies to , which then equals .

The diameter of Circle B is the same as the side of large Square C is . The area is .

The ratio of their areas is then , which reduces to .

To answer this question, you need to find the area of the circle and then subtract out the area of the square. If minor arc AD measures  then the entire circumference is . This is because the arcs formed by the vertices of the square are each exactly  of the circumference. If the circumference is , then the radius is   is how you'd calculate that). This then means that the area of the circle, , is .

So you know everything there is to know about the circle, and now you need to move on to the square. For the square, simply draw in the diameter of the circle and realize that this is also the diameter of the square. Since the diagonal of any square is equal to a side of the square times , the inverse is also true: the side of the square can be calculated as the diagonal divided by . This the side of this square is equal to , which can be simplified to .

The area of the square is side-squared, so that's . So your answer here will be the area of the circle, , minus the area of the square, .  is approximately , so the answer is approximately .

The key here is to recognize that the diameter of the circle () is the length of the sides of the square. We can calculate the area of the square to be the square of the sides, , and the area of the circle to be . The percent of the square’s area taken up by the circle would be the part (circle) over the whole (square) times : , which simplifies to .

The correct response is . To find the circumference of a circle, we must find the radius. Let’s start with what we know:

The area of a rectangle is . Here we are told that . That means the length is . We also know that from X to , so the width of the rectangle must be . Let’s re-draw the shape:

We can see that a right triangle is formed with the radius AX and half the length and width of the rectangle. Since it’s a classic Pythagorean triplet (), we don’t need to use the Pythagorean theorem!

Now that we have the radius, we can plug it into the formula for circumference: . . The circumference is approximately , or a number slightly larger than .

This problem demonstrates several important properties of a circle. Most importantly, you should recognize that all radii are the same so AC = BC = CD. This fact, while fairly simple retrospectively, is easily overlooked when analyzing a circle diagram such as this. For statement I, because all radii are the same, you should note that triangle ACB is isosceles, and thus angle ABC also equals 30° and ACB = 120°. This means that angle BCD = 60° (it is supplementary to angle ACB) and this is the central angle cutting out minor arc BD. As a result, minor arc BD must be  or  of the circumference, which is  . Minor arc BD is thus  and since  is approximately 3.14 you know that 3(3.14) is less than 10 and this is not a true statement. This statement cannot be inferred from the diagram.

Statement II tests another important property of circles. If a triangle is inscribed in a circle with one side as the diameter, the opposite angle in the triangle is always 90°. This is because inscribed angles that cut out a certain arc (those drawn from a point on the circumference) are always equal to half of the central angle cutting out the same arc. If one side is the diameter, then the opposite angle is always cutting out exactly half of the circle. Since that central angle is 180°, the angle in question will always be 90°. This statement must be true based upon the diagram.

Statement III is relatively straightforward if you figured out statement I. In the first statement, you proved that angle BCD must be 60°. Since BC = BD then the other two angles must also be 60° and triangle BCD must be equilateral.

Because statements II and III must be true, but I is not true, the correct answer is II and III only.

Problems with inscribed circles and squares hinge on one important concept: circles and squares are perfectly symmetrical, so if you know one thing about them you tend to know everything. Here you're given the area of a square and asked to find the area of a circle. With the area of a square, you can find out its side lengths, its perimeter - really anything else about it. Since the area of the square is 32, you can quickly take the square root of 32 to find the side length: .

Now you need to find the relationship between the dimensions of the square and the dimensions of the circle. To find the area of a circle you need the length of the radius, which is half the length of the diameter.  And if you look at the diagram, you should see that you can find the diameter of the circle by connecting the corners of the square:

And you know the length of that diagonal: the diagonal of a square is equal to one of its sides times , so you can multiply  to get  as the diameter. The area of a circle formula,  requires the radius, which is half the diameter, so you can plug in  to get an area of  square inches.

When a circle is inscribed in a square as you can see here, the diameter of the circle is the same length as one side of the square. So if you know the area of one shape, you can use that to find the other: here you'll take the area of the square and use that to find the length of a side, then because that is the diameter you can use that to find the radius and apply that to the circle area formula. Your steps should be:

1) Take the square root of the area of the square to find the length of a side. , so the length of one side of the square is . Note that this is also the length of the diameter of the circle!

2) Knowing that the diameter of the circle is , cut that in half so that you have the radius. Half of  is , so the radius here is  and you're ready to plug this into the area formula.

3) Use the area of a circle formula, , to solve for area. That gives you .