# Circles and Conics

Circles and Conics questions on the ACT are straightforward and formula driven. The following chapters will list all the formulas you need to know to answer the circle questions on the ACT Math section.

The formula for finding the circumference of a circle is:

$$\left \{ Circumference = 2\cdot \pi \cdot radius \right \}$$

To find the AREA of a circle use the following formula:

$$\left \{ area = \pi\cdot radius^2\right \}$$

#### ‍EXAMPLE 1:

Points X and Y are two different points on a circle. Point M is located so that line segment XM and line segment YM have equal length. Which of the following could be true?

I. M is the center of the circle

II. M is on arc XY

III. M is outside of the circle

A. I only

B. II only

C. I and II only

D. I, II, and III

Solution :

Option I considers the possibility that M could be the center of the circle if lines XM and YM are equal.. We know this must be true because M being the center point of the circle would make lines XM and YM radii of the circle, which would mean that they are equal.

Option II presents us with the possibility that point M lies somewhere on the arc of XY. If point M rested exactly halfway between X and Y, then straight lines drawn from

X to M and Y to M would certainly be equal.

Option III presents us with the possibility that M lies somewhere on the outside of the circle. So long as M lies at a distance halfway between X and Y, this scenario would still work. So option III is also correct.

This means that all of our options (I, II, and III) are possible. The correct answer is D.

## Revolutions

A full revolution, or turn, of a spinning wheel is equivalent to the wheel going around once. A point on the edge of the wheel travels one circumference in one revolution.

For EXAMPLE, if a wheel spins at 3 revolutions per second, a point on the edge travels a distance equal to 3 circumferences per second. If the wheel has a diameter of 4 feet, then the point travels at a rate of 3 x 4π = 12π feet per second.

#### EXAMPLE 2 :

Sophia is riding her bicycle across a basketball court. She knows her bicycle wheel has a radius of 1 ft and that the basketball court is about 94 ft long. Approximately how many times does her bicycle wheel rotate over the course of her ride across the basketball court?

A. 28 times

B. 30 times

C. 36 times

D. 45 times

Solution: Each time Sophia’s bicycle wheel makes a full rotation, she travels a linear distance of one circumference across the court. The circumference of her wheel is

$$C=d\pi=2r\pi=2(1)\pi=2\pi$$

Now divide the length of the entire basketball court by the diameter of the wheel in order to find the number of revolutions/turns/circumferences Sophia’s wheel must complete to get her to the other side.

$$\frac{94 ft}{2\pi ft/rev}= 30 revolutions$$

## Arc Length

Often, the ACT will ask you to solve for a portion of the distance on a circle, instead of the entire circumference. This portion is termed an ARC.

EXAMPLE: What is the length of arc AXB?

Arc AXB is the arc from A to B, passing through the point X. To find its length, first find the circumference of the circle.

$$C=2\pi r= 24\pi$$

Since the arc is defined by the central angle of 60°, and the entire circle is 360°, then the arc
$$\frac{60}{360}$$=$$\frac{1}{6}$$ of the circle

Therefore, the measure of arc $$AXB=\frac{1}{6}\ (24\pi$$= 4\pi\)

#### Example 3:‍

In the figure above, square RSTU is inscribed in the circle. What is the degree measure of arc ST?

A. 45

B. 60

C. 90

D. 120

Solution: If you draw a line connecting points R and T,you will have a perfect semi-circle, or 180°. .Now, the arc we are looking for spans

exactly half of that semi-circle. This means that the arc degree measure of ST is 900.

## Perimeter of a Sector

If you know the length of the radius and the central (or inscribed) angle, you can find the perimeter of the sector.

#### EXAMPLE 4:

In the figure above, the smaller circles each have a radius of 3. They are tangent to the larger circle at Points A and C and are tangent to each other at point B, which is the center of the larger circle. What is the perimeter of the shaded region?

A.

B.

C. 12π

D. 15π

Solution: Let us start with the two circles in the middle. We know that each circle has a radius of 3 and that our shaded perimeter spans exactly half of each circle.

So the circumference for each small circle is:

C = πr = 3π

And there are two small circles, so we must double this number:

2 x 3π = 6π

So the interior perimeter is 6π.

Now, let’s find the outer perimeter, which is the circumference for half the larger circle. If the radius of each of the small circles is 3, then that means the diameter of

each small circle is 6.This means that the full circumference of the larger semicircle is:

C = πr = 6π

Our outer perimeter equals 6π and our inner perimeter equals 6π. To get the full perimeter, we must add them together.

6π + 6π = 12π

The correct answer is D, 12π

#### EXAMPLE 5:

The two semicircles in the figure above have the centers at R and S respectively. If RS = 12,what is the total length of the darkened curve?

A. 8 π

B. 9 π

C. 12 π

D. 15 π

Solution: Here, we could simply imagine that line RS is the diameter of a complete circle. (Why are we allowed to do this? Because we have the sum of two radii and two half circles, so combined, they would become one circle.) If RS is a diameter of a circle whose complete circumference we must find, let us use our circumference formula.

c = 2πr = 12π

The correct answer is C, 12π

#### EXAMPLE 6:

Point O is the center of both the circles in the figure above.

If the circumference of the larger circle is 36 and the radius of the small circle is half the radius of the larger circle, what is the length of the darkened arc?

A. 10

B. 8

C. 6

D. 4

Solution: We are trying to find the length of an arc circumference, which means that we need two pieces of information - the arc degree measure and the radius.

Well, we already have the degree measure, so we’re halfway there, but now we need the radius(or diameter) of the smaller circle.

We are told that it is half the radius of the larger circle, so we must find the radius of the larger circle first.

All that we are told about the larger circle is that it has a circumference of 36. Luckily, we can find its radius from its circumference.

C = 2πr

36 = 2πr

$$r=\frac{18}{\pi}$$

Because we know that the smaller circle has a radius that is half the length of the radius of the larger circle, we know that the radius of the smaller circle is:

$$r=\frac{1}{2}\times \frac{18}{\pi}=\frac{9}{\pi}$$

$$C_{arc}=2\pi{}r\frac{(arcdegree)}{360^{\circ}}$$

$$C_{arc}=2\pi{}\frac{9}{\pi}\frac{(80^{\circ})}{360^{\circ}}=4$$

## Area of a sector

Often, the ACT will ask you to solve for the area of a sector of a circle, instead of the area of the entire circle. You can find the area of a sector by determining the fraction of the entire area that the sector occupies. To determine this fraction, look at the central angle that defines the sector.

EXAMPLE: What is the area of sector ACB (the striped region) below?

First, find the area of the entire circle:

$$A=\pi{r^2}=\pi(3^2)=9\pi$$

Then, use the central angle to determine what fraction of the entire circle is represented by the sector. Since the sector is defined by the central angle of 60, and the

entire circle is 360°, the sector occupies 60°/360°, or one-sixth, of the area of the circle.

Therefore, the area of sector ACB = (9π) = 1.5π

#### EXAMPLE 7:

Dave claims to have only eaten 120°, of a circular apple pie that has a radius of 7.8 cm. When looking at the leftover pie from the top, it is found that 23.14π $$cm^2$$ of

the flat plate underneath the pie is exposed in the area that Dave ate.

Did Dave truly eat 120°, of the pie?

A. Yes, he ate exactly 120º, of the pie.

B. No, he ate less than 120º of the pie.

C. No, he ate more than 120º of the pie.

D. No, he ate exactly 120º of the pie

Solution: The question asks us to find the proportion of the pie that Dave ate - the area of a sector. Since we are given radius, we can find the area of the circle.

$$A_{pie}=\pi\times (7.8cm)^2$$

$$A_{pie}=60.84\pi\ cm^2$$

Now, we must divide the area of pie that Dave ate (which is given directly in the question) by the area of the whole pie to determine the proportion of pie he really ate

$$\frac{23.14\pi}{60.84\pi}=0.38$$

0.38 of a pie will translate to = 0.38 * 360º = 136.8º.

Therefore, Dave actually ate more than 120º of the pie but not 360º of the pie.

#### EXAMPLE 8:

The circle has an area of 25π and is divided into 8 congruent regions. What is the perimeter of one of these regions?

A. 10 - 25 π

B. 10 + 0.5 π

C. 10 + 1.25 π

D. 10 + 5 π

Solution: The bigger circle has an area of 25π.

$$Area=\pi r^2=25\pi$$

r=5

Now, we must find the arc measurement of each wedge. To do so, let us find the full circumference measurement and divide by the number of wedges (in this case, 8).

The full circumference is C = 2πr = 10π which, divided by 8, is:

$$Arc_{wedge}=\frac{10\pi}{8}=\frac{5\pi}{4}$$

Now, let us add that arc measurement to twice the radius value of the circle in order to get the full perimeter of one of the wedges

$$5+5+\frac{5\pi}{4}$$

$$10+\frac{5\pi}{4}$$

So the correct answer is C.

#### EXAMPLE 9:

In the figure above, triangle ABC is inscribed in the circle with center O and diameter AC. If AB = AO, what is the degree measure of ∠AOB?

A. 15°

B. 30°

C. 45°

D. 60°

Solution: We are told that lines AO and AO are equal.

Based on our knowledge of circles, we also know that AO and AO are equal.This means that AB = AO = BO, which means that the triangle is equilateral. So, the angle measure of ABO = 60°.

So the correct answer is D.

## Clock Math

The ACT may ask you to apply your knowledge of circle geometry to situations specific to clocks. Don’t let these questions throw you off: at this point, your knowledge of circles includes everything you need to solve them, even if they seem somewhat unfamiliar.

#### EXAMPLE 10:

What is the measure, in degrees, of the acute angle formed by the hands of a twelve-hour clock that reads exactly 3:10?

A. 30°

B. 35°

C. 40°

D. 45°

Solution: One hand is pointing at the 2, while the other is pointing at the 3, so they’re 1/12 of a 360° circle apart (there are 12 numbers on a typical clock),

right? = 360°/12 = 30, so the answer must be 30, right?

Not so fast. One has to account for the 10 minutes that have passed. Ten minutes is 1/6 of an hour, so the hour hand has also moved 1/6 of the distance between the 3

and the 4, which adds 5 (1/6 of 30). The total measure of the angle, therefore, is 35, and the correct answer is B.

#### EXAMPLE 11:

On a standard analog clock, what is the angle between the hands when the clock reads 11:20?Give the smaller of the two angles.

A. 60°

B. 70°

C. 75°

D. 80°

Solution: To find the degrees of a clock hand, first find the angle between each hour-long sections. As calculated for the last question, the angle between each hour is

60°.

At 11:20, the hour hand has gone one-third of the way between the 11 and the 12. Thus, there are two-thirds of 30° between the hour hand and the 12. (2/3) 30° = 20°.

There are 60° between 12 and 2, where the minute hands is, which means there’s a total of 20° + 60° = 80° between the hands.

## Equation of a Circle

The standard equation for a circle is in the form $$(x-h)^2 + (y-k)^2 = r^2$$, where the circle with radius r has its center at (h, k). Let’s consider an EXAMPLE:

#### EXAMPLE 12:

In which quadrant(s) will the circle with equation $$(x + 8)^2 + (y-15)^2 = 49$$ be contained?

A. I only

B. II only

C. II and III only

D. IV only

Solution: Begin by finding the center of the circle. The center of the circle is at (-8, 15), which is in Quadrant II. The radius of this circle is 7.

Now, we need to consider the four points seven units in the positive/negative x-direction and positive/negative y-direction. These will be on the line of the circle and

mark the farthest points in those directions. If all of those points are located in Quadrant II, we know that the entire circle is located in that quadrant.

Let’s consider those points now:

All those points are in Quadrant II, and B is the correct answer.

#### EXAMPLE 13 :

Find the centre and radius of the circle $$x^2 + y^2 + 8x + 7 = 0$$ .

A. Centre is (-4,0) and radius is 9

B. Centre is (0,-4) and radius is 9

C. Centre is (-4,0) and radius is 3

D. Centre is (0,-4) and radius is 3

Solution: There is no Y term, so we have to complete the squares only for the X terms

$$x^2 + 8x +\; \: \: \: \: + y^2=-7$$

We need to fill in the ___ to complete the square. The middle term is 8 and therefore the last

variable should be 16. Adding 16 on both sides:

$$x^2+ 8x + 16 + y^2= -7 + 16$$

$$(x + 4)^2+ y^2= 9$$

Comparing this equation to the equation of a circle: $$(x-h)^2 + (y-k)^2 = r^2$$

We get (h,k) -> (-4,0) and r = 3.

# Conics

A CONIC SECTION (or simply conic) can be described as the intersection of a plane and a double-napped cone

Notice from the above figure, the formation of the four basic conics:

1. Circle

2. Ellipse

3. Parabola

4. Hyperbola

Thankfully, conics are rarely ever tested on the ACT. Even when they are, the questions are extremely straightforward and generally

only require the applications of the basic formulae.Therefore, the information we present in this chapter are the very basics and nothing

more.

## 1. Parabola

The shape of a parabola is shown in this picture.

Notice that the parabola a line of symmetry,meaning the two sides mirror each other.

There are two patterns for a parabola, as it can be either vertical (opens up or down) or horizontal (opens left or right.)

Let's look at a few key points about these patterns:

• If the x is squared, the parabola is vertical (opens up or down).
• If the y is squared, it is horizontal (opens left or right).
• If a is positive, the parabola opens up or to the right.
• If a is negative, it opens down or to the left.

The vertex is at (h, k). You have to be very careful. Notice how the location of h and k switches based on if the parabola is vertical or horizontal. Also, the coordinate

inside the parenthesis is negative, but the one outside is not.

Let's look at a couple parabolas and see what we can determine about them.

$$\textbf{1.y = -2(x+3)}^2\textbf{+4}$$

First, we know that this parabola is vertical (opens either up or down) because the x is squared.We can determine it opens down because the a (-2) is negative.

Next we can find the vertex (h, k). For a vertical parabola, h is inside parenthesis, and since there is a negative in the pattern, we must take the opposite. So h = -3. k is

outside, and the sign is positive, so we will keep this number as is. k = 4. Thus, our vertex is (-3, 4).

Summary: This is a vertical parabola that opens down. Its vertex is (-3, 4).

EXAMPLE: Write $$y=3x^2+24x+50$$ in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.

$$y=3x^2+24x+50$$                                     Original equation

$$y=3(x^2+8x)+50$$                                      Factor 3 from x-terms

$$y=3(x^2+8x+(\ \ ))+50-3(\ \ )$$                 Complete the square on the right side

$$y=3(x^2+8x+(16))+50-3(16)$$                  The 16 added when you complete the square is multiplied by 3

$$y = 3(x + 4)2 + 2$$

$$y = 3[x - (-4)]2 + 2$$                                   (h, k) = (-4, 2)

The vertex of this parabola is located at (-4, 2), and the equation of the symmetry is x = -4.The parabola opens upward.

GRAPH PARABOLAS: You can use symmetry and translations to graph parabolas. The equation $$y = a(x - h)^2 + k$$ can be obtained from $$y = ax^2$$ by replacing x with x - h and y with y - k.

Therefore, the graph of $$y = a(x - h)^2 + k$$ is the graph of $$y = ax^2$$ translated h units to the right and k units up.

Graph each equation.

$$\textbf{1. y = -2x}^2$$

For this equation, h = 0 and k = 0.The vertex is at the origin. Since the equation of the axis of symmetry is x = 0, substitute some

small positive integers for x and find the corresponding y-values.

Since the graph is symmetric about the y-axis, the points at

(-1, -2), (-2, -8), and (-3, -18) are also on the parabola.

Use all of these points to draw the graph.

$$\textbf{2. y = -2(x-2)}^2 + \textbf{3}^2$$

The equation is of the form $$\textbf{y=a(x-h)}^{2}+\textbf{k}$$ ,where h = 2 and k = 3. The graph of this equation is the graph of

$$\textbf{y=-2x}^\textbf{2}$$ in part a translated 2 units to the right and up 3 units. The vertex is now at (2, 3).

## 2. Ellipse

There are two patterns for an ellipse:

In both patterns, (h, k) is the center point, just as it was with a circle. The a and the b have to do with how wide and how tall the ellipse is.

Each ellipse has a major axis and a minor axis. The major axis is the line that goes through the longest length of the ellipse. The minor axis goes through the shorter

distance.

The major axis is always associated with the variable a and equals 2a. The minor axis is always associated with b and

equals 2b.

You'll notice there are two different patterns: one for a horizontal ellipse and one for a vertical.Whichever denominator

is larger determines which variable is a (because a is always bigger since it is the major axis.) If the larger number is

under the x, then the ellipse is horizontal. If it is under the y then it is vertical.

EXAMPLE:   $$\frac{(x-2)^2}{16}+\frac{(y-1)^2}{9}=1$$

Let's start by finding the center point, which is (2, 1). Now we need to decide if this ellipse is vertical or horizontal. 16 is

larger than 9, so that means 16 is $$a^2$$ (since it is larger). Whenever the larger number ($$a^2$$) is underneath the x,

then it is a horizontal ellipse. (You can remember this because the x-axis is the horizontal axis.)

We want to identify one last piece of information: the length of a and b. Notice in the pattern that both of these are squared, so we have to take the square root of each.

Thus, a = 4 and b = 3 This is a horizontal ellipse with a center point of (2, 1).

Each ellipse has two foci (plural of focus) as shown in the picture here:

As you can see, c is the distance from the center to a focus We can find the value of c by using the formula  $$c^2=a^2-b^2$$

EXAMPLE: Find the centre and foci for the following ellipse?

$$\frac{(x-4)^2}{25}+\frac{(y-1)^2}{9}=1$$

This ellipse has already been graphed and its center point is marked. The centre is (4,1)

We need to use the formula $$c^2=a^2-b^2$$ to find c.

$$c^2=a^2-b^2$$

$$c^2=25-9$$

c=4

We can plot the foci by counting 4 spaces from the center. Since the ellipse is horizontal, we will count 4 spaces left and right and

plot the foci there. The foci is (4+4,1) and(4-4,1) = (8,1) and (0,1) are the foci (red dots in the diagram)of the above parabola.

## 3. Hyperbola

There are two patterns for hyperbolas

By examining the equation, we can determine the following:

1. If it is vertical or horizontal.If the x term is positive, the parabola is horizontal (the curves open left and right). If the y term is positive, the parabola is vertical (the curves open up and down). Unlike an ellipse, it does not matter which denominator is larger.

2. The center point.As with all conic sections, the center point is (h, k). Notice that the h is always with the x and the k is always with the y. There is also a negative in front of each, so you must take the opposite.

3. The a and b values.These will be needed to graph the parabola. Notice that a is always under the positive term and b is always under the negative term.

EXAMPLE: Let's identify key information from this hyperbolas:

$$\frac{(x-3)^2}{9}+\frac{(y-2)^2}{25}=1$$

First of all, we know i

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