Circles - ACT Math

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Question

What is the area of a circle with a circumference of $12\pi$ ?

Answer

This problem tests your ability to work with the two most important circle formulas (both given to you in your SAT test booklet). The circumference of a circle is $C=2\pi r$ , and the area of a circle is $A=\pi r^2$ . Since you’re given the circumference of the circle as $12\pi$ , you can set up an equation to solve for , the radius:

$12 \pi =2\pi r$

Then you can plug in 6 as the radius in the area formula:

$A=\pi (6^2)=36\pi$

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Question

A circle has an area of $144\pi$ . What is its circumference?

Answer

This problem tests your ability to work with the two most important circle formulas (both given to you in your SAT test booklet). The circumference of a circle is $C=2\pi r$ , and the area of a circle is $A=\pi r^2$ . Since you’re given the area of the circle as $144\pi$ , you can set up an equation to solve for , the radius:

$144\pi =\pi r^2$

Now you can plug in that radius of 12 into the circumference formula:

$C=2\pi r$

$C=2\pi (12)$

$C=24\pi$

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Question

If the circle above has center A and area $144\pi$ , what is the perimeter of sector ABCD?

Answer

This problem tests several of the core properties of circles:

Area = $\pi r^2$

Circumference = $\pi d = 2\pi r$

Arc length = $\frac{central angle}{360}\times circumference$

Here you're given the area, but to determine the perimeter of that sector you need to find the radius (for line segments AB and AD) and the arc length BCD. With an area of $144\pi$ that means that the radius is 12. Since you need two radii (AB and AD) to form the "legs" of the sector, that means that the straight-line legs sum to 24 (a good hint that your correct answer should include the number 24).

For the arc length, note that the central angle measures 45, and that $\frac{45}{360}=\frac{1}{8}$ . So the arc BD will equal one-eighth of the circumference. The circumference is $2\pi r=24\pi$ , so you can find the arc by calculating $\frac{1}{8}(24\pi )=3\pi$ . Adding together the two legs plus the arc, you get your answer $24+3\pi$

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Question

The area of the semicircle above is $32\pi$ . What is the length of the arc connecting points A and B?

Answer

This problem puts a slight twist on the area of a circle formula, $A=\pi r^2$ . You’re given the area of half a circle as $32\pi$ , so to apply the area formula you should double $32\pi$ to account for the missing half of the circle so you can use the classic area formula. Since the area of the entire circle would be $64\pi$ 64pi, you can use the area formula to solve for the radius:

$64\pi =\pi r^2$

So the radius is 8, which means you’re ready to apply the formula for circumference of a circle. But again, you’re only dealing with half the circumference since the problem asks for the arc length of the semicircle, so while the formula for circumference i s $C=2\pi r$ , you can cut that in half to find the distance of half the circumference. That means you need to calculate $\pi r$ . With a radius of 8, that means that your answer is $8\pi$ .

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Question

A pizzeria measures its pizzas by their diameters when listing sizes on the menu. What is the difference, in square inches, in surface area between a 10-inch pizza and an 8-inch pizza from that pizzeria (assume that each pizza has a negligible height/thickness)?

Answer

An extremely important consideration on this problem is that the sizes are quoted in terms of the length of the diameter, but the area formula requires you to work with the radius. So as you calculate the area of each pizza, it is very important to first divide the diameter by 2 so that you are working with the radius in the formula $A=\pi r^2$ .

10-inch pizza → radius of 5 → $A=\pi (5^2)=25\pi$ square inches

8-inch pizza → radius of 4 → $A=\pi (4^2)=16\pi$ square inches

Therefore the difference in surface area is $25\pi -16\pi =9\pi$ square inches.

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Question

In the circle above, centered on point O, angle AOB measures 40 degrees. If line segment BD measures 18 inches, what is the measure, in inches, of minor arc BC?

Answer

To calculate an arc length, you need two pieces of information:

The circumference of the circle (calculated as $2\pi r$ or $\pi d$ , where is the radius and is the diameter)

The measure of the central angle that connects the two end points of the arc.

Then the arc length is the proportion of the circumference represented by that angle: multiply $\frac{angle}{360}$ by the circumference and you have the arc length.

Here you're given the diameter (via line segment BD) as 18, so you know that the circumference is $18\pi$ . And you know that angle AOB measures 40 degrees, so you can conclude that angle BOC is 140 degrees, since angles AOB and BOC must complete the straight line AC and straight lines measure 180 degrees.

Therefore your calculation is $\frac{140}{360}(18\pi )$ , which simplifies to $\frac{7}{18}(18\pi)=7\pi$ .

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Question

A bicycle tire has a diameter of 70 centimeters. Approximately how many revolutions does the tire make if the bicycle travels 1 kilometer? (1 kilometer = 1000 meters = 100000 centimeters)

Answer

The distance around the outside of a circle is, of course, the circumference. A common way to test the circumference in a word problem is to use the circumference of a wheel as a straight line distance: for each revolution of the wheel, a vehicle will travel the length of the circumference.

Here you know that the diameter of the wheel is 70 centimeters, which means that the circumference, calculated as $\pi d$ , is $70\pi$ . Note that the question asks for "approximately" the number of revolutions, and that the answer choices are spread quite far apart. This means that you can use an estimate of 3.14 or $\frac{22}{7}$ for $\pi$ and say that one revolution moves the wheel approximately 220 centimeters. Since the wheel needs to cover 100000 centimeters, you should then divide 100000 centimeters by 220 to see that the answer is approximately 450.

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Question

In the figure above, AC is the diameter of a circle with center O. If the area of the circle is $81\pi$ , what is the length of minor arc BC?

Answer

To calculate an arc length, such as the length of minor arc BC here, your job is to find the proportion that that arc represents out of the total circumference. So you'll need to find 1) the circumference and 2) the measure of the central angle of that arc.

Here since you know that the area is $81\pi$ , you can use the formula $Area =\pi r^2$ to determine that the radius measures 9.

Then you can plug in that radius into the circumference formula, $Circumference = 2\pi r$ , to find that the circumference measures $18\pi .$

From this point, you need to find the measure of angle BOC. Since angle AOC measures 180 degrees (you know that it's a straight line, because it's defined as a diameter), and angle AOB = 110, that means that BOC measures 70 degrees. So minor arc BC will be $\frac{70}{360}$ of the total circumference, setting up the calculation:

$\frac{70}{360}(18\pi )$

That reduces to $\frac{7}{2}\pi$ which equals $3.5\pi$ .

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Question

A gas pipe has an outside diameter of 24 inches. The steel wall of the pipe is 1 inch thick. What is the area of the cross-section of the steel wall of the pipe?

Answer

The cross section can be found by calculating the area of the larger, diameter 24, circle (inclusive of the pipe in gray) and subtracting the area of the smaller, diameter 22, circle (everything in white inside of the pipe). To do this, use the area formula: $Area = \pi r^2$ .

The radius of the larger circle is 12 and of the smaller circle is 11, meaning that your areas can be calculated as:

Outer/Larger Circle: $\pi (12^2)=144\pi$

Inner/Smaller Circle: $\pi (11^2)=121\pi$

Then subtract the areas and you'll be left with just the area of the gray ring: $144\pi -121\pi =23\pi$

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Question

On a certain circular dartboard, the diameter of the circular bullseye is 4 inches and the diameter of the entire board is 20 inches. If a dart hits the board in a random location, what is the probability that it hits the bullseye target?

Answer

While this problem asks about probability, it's essentially an area of a circle problem. The probability of hitting the bullseye is just the area of the bullseye divided by the area of the dartboard in total.

Since $Area=\pi r^2$ your job becomes to find the radius of each circle. You're quoted the diameters, so to find the radius just divide the diameter by 2.

Bullseye: Diameter = 4 so Radius = 2. Area = $\pi (2^2)=4\pi$

Dartboard: Diameter = 20 so Radius = 10. Area = $\pi (10^2)=100\pi$

So the probability of hitting the bullseye is $\frac{4\pi }{100\pi }$ which reduces to $\frac{1}{25}$ .

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Question

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?

Answer

The formula for the equation of a circle is:

$(x-h)^{2}+(y-k)^{2}=r^{2}$

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

and

and diameter = 6 therefore radius = 3

$(x-0)^{2}+(y-4)^{2}=3^{2}$

$x^{2}+(y-4)^{2}=9$

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Question

Circle A is given by the equation $(x-4)^{2}+(y+3)^{2}=29$ . 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?

Answer

The general equation of a circle is $(x-h)^{2}+(y-k)^{2}=r^{2}$ , 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 $(x-4)^{2}+(y+3)^{2}=29$ . This means that its center must be located at (4, –3), and its radius is $\sqrt{29}$ .

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 $2\sqrt{29}$ .

Now, that we have circle A's new center and radius, we can write its general equation using $(x-h)^{2}+(y-k)^{2}=r^{2}$ .

$(x-(-2))^{2}+(y-2)^{2}=(2\sqrt{29})^{2}=2^{2}(\sqrt{29})^{2}=4(29)=116$ .

$(x+2)^{2}+(y-2)^{2}=116$ .

The answer is $(x+2)^{2}+(y-2)^{2}=116$ .

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Question

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

$y_{1}=4x+3$

and

$y_{2}=5x+44$ ?

Answer

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 $y_{1}$ :

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

Now, recall that the general form for a circle with center at $(x_{0},y_{0})$ is:

$(x-x_{0})^{2}+(y-y_{0})^{2}=r^{2}$

For our data, this means that our equation is:

$(x+41)^{2}+(y+161)^{2}=12^{2}$ or $(x+41)^{2}+(y+161)^{2}=144$

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Question

What is the radius of a circle with the equation ?

Answer

We need to expand this equation to $x^{2}-8x+y^{2}-6y=24$ and then complete the square.

This brings us to $x^{2}-8x+16+y^{2}-6y+9=24+16+9$ .

We simplify this to $(x-4)^{2}+(y-3)^{2}=49$ .

Thus the radius is 7.

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Question

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

Answer

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

$r=\sqrt{74}$

This radical cannot be reduced further.

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Question

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?

Answer

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

$x^{2}+y^{2}=r^{2}$

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

$x^{2}+y^{2}=5^{2}$

$x^{2}+y^{2}=25$

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

$2^{2}+y^{2}=25$

$4+y^{2}=25$

$y^{2}=21$

$y=\pm \sqrt{(21)}$

Since our answer will be positive, it must be $\sqrt{(21)}$ .

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Question

The following circle is moved $\frac{\pi}{2}$ spaces to the left. Where is its new center located?

$(x-\frac{\pi}{2})^2 + (y - \frac{3\pi}{2})^2 = \pi$

Answer

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

With that in mind, our original center is at $(\frac{\pi}{2}, \frac{3\pi}{2})$ .

If we move the center $\frac{\pi}{2}$ units to the left, that means that we are subtracting $\frac{\pi}{2}$ from our given coordinates.

$\frac{\pi}{2} - \frac{\pi}{2} = 0$

$\frac{3\pi}{2} - \frac{\pi}{2} = \frac{2\pi}{2} = \pi$

Therefore, our new center is $(0, \pi)$ .

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Question

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

Answer

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

$(x-h)^{2} + (y-k)^{2} = r ^{2}$

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 $x_{1} = y_{1} = 0 , x_{2} = y_{2} = 9$ in the midpoint formula:

$h =\frac{ x_{1}+ x_{2} }{2} = \frac{0+ 9}{2} = \frac{9}{2}$

and

$k=\frac{ y_{1}+ y_{2} }{2} = \frac{0+ 9}{2} = \frac{9}{2}$

The center of the circumscribing circle is therefore $\left ( \frac{9}{2}, \frac{9}{2} \right )$ .

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:

$r= \sqrt{(x_{2} - x_{1})^{2} +(y_{2} - y_{1})^{2} }$

Since we are actually trying to find $r ^{2}$ , we will use the form

$r^{2}= (x_{2} - x_{1})^{2} +(y_{2} - y_{1})^{2}$

Choosing the radius with endpoints and $\left ( \frac{9}{2}, \frac{9}{2} \right )$ , we set $x_{1} = y_{1} = 0 , x_{2} = y_{2} = \frac{9}{2}$ and substitute:

$r^{2}= \left ( \frac{9}{2} - 0 \right ) ^{2} + \left ( \frac{9}{2} - 0 \right ) ^{2}$

$= \left ( \frac{9}{2} \right )^{2} + \left ( \frac{9}{2} \right )^{2}$

$= \frac{81}{4 } + \frac{81}{4 }$

$= \frac{81}{2}$

Setting $h =k= \frac{9}{2}$ and $r^{2}= \frac{81}{2}$ and substituting in the circle equation:

$(x-h)^{2} + (y-k)^{2} = r ^{2}$

$\left ( x - \frac{9}{2}\right )^{2} + \left ( y- \frac{9}{2}\right )^{2} = \frac{81}{2}$ , the correct response.

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Question

The above figure shows a circle on the coordinate axes with its center at the origin. $\overarc {AB}$ has length $6 \pi$ .

Give the equation of the circle.

Answer

A $135^{\circ }$ arc of a circle represents $\frac{135}{360 } = \frac{3}{8}$ of the circle, so the length of the arc is three-eighths its circumference. Set up the equation and solve for :

$\frac{3}{8}C = 6 \pi$

$\frac{8} {3} \cdot \frac{3}{8}C =\frac{8} {3} \cdot 6 \pi$

$C = 16 \pi$

The equation of a circle on the coordinate plane is

$(x- h)^{2}+ (y - k)^2 = r^{2}$ ,

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

The radius of a circle can be determined by dividing its circumference by $2 \pi$ , so

$r = \frac{C}{2 \pi } = \frac{16 \pi }{2 \pi } = 8$

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

$(x- 0)^{2}+ (y - 0)^2 =8^{2}$ ,

or

$x^{2}+ y^2 =64$ .

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Question

The above figure shows a circle on the coordinate axes with its center at the origin. The shaded region has area $49 \pi$ .

Give the equation of the circle.

Answer

The unshaded region is a $45 ^{\circ }$ sector of the circle, making the shaded region a $(360 - 45) ^{\circ } = 305 ^{\circ }$ sector, which represents $\frac{305}{360} = \frac{7}{8}$ of the circle. Therefore, if is the area of the circle, the area of the sector is $\frac{7}{8}A$ . The sector has area $49 \pi$ , so

$\frac{7}{8}A = 49 \pi$

Solve for :

$\frac{8} {7} \cdot \frac{7}{8}A =\frac{8} {7} \cdot 49 \pi$

$A =56\pi$

The equation of a circle on the coordinate plane is

$(x- h)^{2}+ (y - k)^2 = r^{2}$ ,

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

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

$A = \pi r ^{2}$ .

Set $A = 56 \pi$ and solve for $r^{2}$ :

$\pi r ^{2} = 56 \pi$

$\frac{\pi r ^{2} }{\pi}=\frac{ 56 \pi}{\pi}$

$r ^{2} = 56$

The center of the circle is , so . Substituting 0, 0, and 56 for , , and $r^{2}$ , respectively, the equation of the circle becomes

$(x- 0)^{2}+ (y - 0)^2 =56$ ,

or

$x^{2}+ y^2 =56$ .

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