The standard form of the equation of a circle is 

,

where the radius is  and the center is .

The equation given is this same form, with  replacing ,  replacing , and  replacing , so to find the radius, we need to find . 

Statement 1 alone tells us that the center is  but it tells us nothing about the radius. Statement 2 alone tells us only that the circle passes through . 

The two together, however, reveal enough information to give the radius. The radius is the distance from the center to a point on the circle, so we can use the distance formula to find the distance between  and . This is the radius.

We are given the area and circumference of a circle and asked to find the radius.

Given the following equations:

We can use either equation to work backwards and find our radius, therefore; Each statement alone is enough to solve the question.

Statement 1) gives the circumference of a circle.  The formula for finding the circumference of a circle is .  The radius can be solved by using this formula.

Statement 2) gives the standard form of a circle, where  is the center of the circle:

The radius is also given in the equation.

Therefore, either statement alone is enough to solve for the radius of a circle.

To find the radius of a cricle, we need its circumference, diameter, or area.

I) Seems to be helpful, but it is really just giving us pi, so it is not sufficient.

II) Gives us the area of the circle, which we can use to work backward to find the radius. 

So II is sufficient to answer the question, but I is not.

The length of a diagonal of a square inscribed inside a circle is equal to the diameter of the circle, and half this is the radius. If a right triangle is inscribed inside a circle, The hypotenuse of a right triangle inscribed inside a circle is also a diameter, and half the length is the radius. Since each statement alone mentions an inscribed square in one circle to an inscribed right triangle in the other, we need only compare the length of a diagonal of the former to that of the hypotenuse of the latter.

Assume Statement 1 alone. The square inscribed inside Circle 2 has perimeter 48; its sidelength is one fourth this, or 12, and the length of a diagonal is  times this, or . The hypotenuse of the right triangle inscribed in Circle 1, with both legs equal to 10 - which is a 45-45-90 triangle, being isosceles - is, by the 45-45-90 Theorem,  times the length of a leg, or . The diagonal of the square inscribed inside Circle 2 is longer than the hypotenuse of the triangle inscribed inside Circle 1, so Circle 2 has the greater radius.

Assume Statement 2 alone. A square of area 100 - and, subsequently, of sidelength the square root of this, or 10 - can be inscribed inside Circle 1; its diagonal will have length  this, or .

The 30-60-90 triangle inscribed inside Circle 2 has a leg of length . However, Statement 2 does not make it clear whether the leg is the shorter leg or the longer leg. If it is the shorter leg, then by the 30-60-90 Theorem, the hypotenuse is twice , or ; if it is the longer leg, then by the 30-60-90 Theorem, the hypotenuse is  times , or 

.

In the first scenario, since , Circle 2 has the longer radius. In the second, since  (this can be seen by noting that  and  ), Circle 1 has the greater radius.

Assume Statement 1 alone. The length of a segment with the given endpoints can be calculated using the distance formula. However, it is not clear whether the points are opposite vertices, in which case the segment is a diagonal of the square, or the points are consecutive vertices, in which case the segment is a side of the square, making the diagonal of the square  times this length. The length of the diagonal of the inscribed square cannot be determined for certain; since the diameter of the circle is equal to the length of the diagonal, the diameter cannot be determined, and since the radius is half this, the radius cannot be determined.

Assume Statement 2 alone. The length of a segment with the endpoints can be calculated using the distance formula. However, it is not clear whether the segment is a hypotenuse of the triangle or not; the diameter of a circle is equal to the length of the hypotenuse of an inscribed right triangle, so knowing this is necessary.

Assume both statements to be true. The two statements together give four points of the circle; since three points uniquely define a circle, the circle can be located; subsequently, the radius can be found.

The diameter of a circle that circumscribes a rectangle is equal to the length of a diagonal of the rectangle; the radius is equal to half this.

The two statements together, however, do not yield this. The opposite sides of a rectangle are congruent, so the two statements are actually equivalent; each gives the same dimension of the rectangle. This is insufficient to determine the length of the diagonal. 

First, locate the other midpoints of the sides of the triangle and construct the segments from each vertex to the opposite midpoint.

Since  is equilateral, , , and  are all altitudes that insersect at the center of the circumscribed circle, , so that .  is the radius of the circumscribed circle.

Assume Statement 1 alone. The length of one side of an equilateral triangle can be calculated using the formula

, or, equivalently,

Once  is calculated, then, since   is also a perpendicular bisector of  and a bisector of , making  a 30-60-90 triangle,  can be calculated to be one half of ;  can be multiplied by  to yield , and, since the three altitudes of an equilateral triangle divide one another into segments whose lengths have ratio 2:1,  can be multiplied by  to obtain radius .

Statement 2 gives us  explicitly, so we can take two thirds of this to get the radius 

.

Assume Statement 1 alone. A  arc of Circle 1 has  the circumference of Circle 1. Since this is also  the circumference of Circle 2, then, if we let  be the circumferences,

,

and

 .

This gives Circle 2 the greater circumference and, subsequently, the greater radius.

Assume Statement 2 alone. A  sector of Circle 2 has area  of Circle 2. Since its area is also equal to  that of Circle 1, then, if  are the areas of Circle 1 and Circle 2, respectively, then 

,

and 

This gives Circle 2 the greater area and, subsequently, the greater radius.

If a right triangle can be inscribed inside a given circle, then its hypotenuse has a length equal to the diameter of the circle, and the radius of the circle can be calculated as half this. Statement 1 gives sufficient information to find this, since the length of the hypotenuse is the distance between its endpoints  and , which is ; the diameter of the circle is 20, and the radius is half this, or 10. From Statement 2, we can only find the length of one leg of an inscribed right triangle, so the length of the hypotenuse is still open to question.