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ACT Math : ACT Math

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Example Questions

Example Question #1 : How To Find The Equation Of A Circle

What is the radius of a circle with the equation x(x-8)+y(y-6)=24 ?

Possible Answers:

Correct answer:

Explanation:

We need to expand this equation to $\dpi{100} \small x^{2}-8x+y^{2}-6y=24$ and then complete the square.

This brings us to $\dpi{100} \small x^{2}-8x+16+y^{2}-6y+9=24+16+9$ .

We simplify this to $\dpi{100} \small \left ( x-4 \right )^{2}+\left ( y-3 \right )^{2}=49$ .

Thus the radius is 7.

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Example Question #1 : Circles

A circle has its origin at (0,0) . The point (5,7) is on the edge of the circle. What is the radius of the circle?

Possible Answers:

$r=2\sqrt{18}$

$r=4\sqrt{21}$

$r=6\sqrt{2}$

There is not enough information to answer this question.

$r=\sqrt{74}$

Correct answer:

$r=\sqrt{74}$

Explanation:

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

r^2=5^2+7^2

r^2=25+49= 74

$r=\sqrt{74}$

This radical cannot be reduced further.

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Example Question #2 : Circles

The endpoints of a diameter of circle A are located at points (-2,4) and (6,8) . What is the area of the circle?

Possible Answers:

$80\pi$

$20\pi$

$40\pi$

$160\pi$

$12\pi$

Correct answer:

$20\pi$

Explanation:

The formula for the area of a circle is given by A =πr². The problem gives us the endpoints of the diameter of the circle. Using the distance formula, we can find the length of the diameter. Then, because we know that the radius (r) is half the length of the diameter, we can find the length of r. Finally, we can use the formula A =πr² to find the area.

The distance formula is

The distance between the endpoints of the diameter of the circle is:

$d=\sqrt{(6-(-2))^{2}+(8-4)^{2}}$

$d=\sqrt{8^{2}+4^{2}}$

$d=\sqrt{64+16}$

$d=\sqrt{80}=4\sqrt{5}$

To find the radius, we divide d (the length of the diameter) by two.

$r=\frac{d}{2}=\frac{4\sqrt{5}}{2}=2\sqrt{5}$

Then we substitute the value of r into the formula for the area of a circle.

$A=\pi r^{2}=\pi ({2\sqrt{5}})^{2}=20\pi$

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Example Question #811 : Act Math

What is the equation for a circle of radius 9, centered at the intersection of the following two lines?

y_1 = 2x + 14

y_2 = 4x + 26

Possible Answers:

(x + 3)^2 + (y - 1)^2 = 18

(x - 7)^2 + (y - 6.5)^2 = 9

(x - 14)^2 + (y - 26)^2 = 9

(x + 6)^2 + (y - 2)^2 = 81

(x - 6)^2 + (y + 2)^2 = 81

Correct answer:

(x + 6)^2 + (y - 2)^2 = 81

Explanation:

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

2x + 14 = 4x + 26

14 = 2x + 26

-12 = 2x

x = -6

To find the y-coordinate, substitute into one of the equations. Let's use y_1 :

$y = 2 \cdot (-6) + 14$

y = -12 + 14

y = 2

The center of our circle is therefore (-6, 2) .

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 - (-6))^2 + (y - 2)^2 = 92

(x + 6)^2 + (y - 2)^2 = 81

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Example Question #11 : How To Find The Equation Of A Circle

What is the equation of a circle with a center of (4,8) and a diameter of ?

Possible Answers:

(x-4)^2+(y+8)^2 = 4

(x-4)^2+(y-8)^2 = 4

(x+4)^2+(y+8)^2 = 4

(x-2)^2+(y-4)^2 = 4

(x-4)^2+(y-8)^2 = 16

Correct answer:

(x-4)^2+(y-8)^2 = 4

Explanation:

Recall that the equation of a circle is defined as:

(x-x_0)^2+(y-y_0)^2 = r^2

Where (x_0,y_0) is the center of the circle.

Given the data we have, we know that the radius of the circle must be (half the diameter). Thus, we know that the equation of the circle in question must be:

(x-4)^2+(y-8)^2 = 4

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Example Question #12 : How To Find The Equation Of A Circle

A circle has a diameter defined by the points (-4,10) and (16,10) . What is the equation of this circle?

Possible Answers:

(x+8)^2+(y-10)^2 = 10

(x+4)^2+(y-6)^2 = 25

(x-6)^2+(y-10)^2 = 400

(x-6)^2+(y-10)^2 = 100

(x+3)^2+(y-5)^2 = 10

Correct answer:

(x-6)^2+(y-10)^2 = 100

Explanation:

Recall that the equation of a circle is defined as:

(x-x_0)^2+(y-y_0)^2 = r^2

Where (x_0,y_0) is the center of the circle.

So, you must first find the center of the circle in question. This you can do by finding the midpoint of the two points given to us. Since they represent a diameter of the circle, their midpoint must be the center of the circle.

Recall that the midpoint of two points is found by the equation:

$(\frac{x_0+x_1}{2},\frac{y_0+y_1}{2})$

Thus, for our points, we have:

$(\frac{-4+16}{2},\frac{10+10}{2})$

This is:

$(\frac{12}{2},\frac{20}{2})$ or (6,10)

Now, the distance between our two points is very easy, as it lies on a horizontal line. Thus, it is just:

16-(-4)=16+4=20

If this is the diameter, the radius of the circle is . Thus, we know based on our data that our circle's equation must be:

(x-6)^2+(y-10)^2 = 100

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Example Question #13 : How To Find The Equation Of A Circle

Which of the following is the equation of a circle with a center at (s,t) with a radius of ?

Possible Answers:

(x + s)^2 + (y+t)^2=400

x^2-2sx+s^2 + y^2 -2ty +t^2 = 400

(x - s)^2 + (y+t)^2=400

2x^2+3sx-s^2 + y^2 -4ty +4t^2 = 400

20x^2+40sx-20s^2 + 20y^2 +40ty -20 t^2 = 0

Correct answer:

x^2-2sx+s^2 + y^2 -2ty +t^2 = 400

Explanation:

To begin, recall that the equation of a circle is defined as:

(x - x_0)^2 + (y-y_0)^2=r^2 , where (x_0,y_0) is your center point.

Now, for this question it is a bit trickier, for our center point is not a pair of numbers but instead is a set of variables (or at least constants that are not specific numbers). So, for this information, you would know:

(x - s)^2 + (y-t)^2=20^2

None of your answers are in this format (except two that are obviously wrong because of the signs). You need to foil out your groups to find the right answer:

(x - s)(x-s) + (y-t)(y-t)=400

Carefully done, this is:

x^2-2sx+s^2 + y^2 -2ty +t^2 = 400

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Example Question #21 : How To Find The Equation Of A Circle

What is the equation of a circle centered around the point (2, 4) with a radius of ?

Possible Answers:

(x-2)^2 + (y-4)^2 = 5

(x-4)^2 + (y-2)^2 = 5^2

(x-2)^2 + (y-4)^2 = 5^2

(x+2)^2 + (y+4)^2 = 5^2

x^2 + y^2 = 5^2

Correct answer:

(x-2)^2 + (y-4)^2 = 5^2

Explanation:

The formula for a circle centered around a point (a,b) with radius is given by:

(x-a)^2 + (y-b)^2 = r^2 .

Thus we see the answer is

(x-2)^2 + (y-4)^2 = 5^2

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Example Question #21 : How To Find The Equation Of A Circle

What is the equation of a circle centered about the origin with a radius of 7? Simplify all exponential expressions if possible.

Possible Answers:

x^2-y^2=49

x^2+y^2=7

x^2 + y^2 = 49

(x-7)^2 + (y-7)^2 = 49

x+y = 49

Correct answer:

x^2 + y^2 = 49

Explanation:

The general formula for a circle centered about points (h,k) with a radius of is:

(x-h)^2+(y-k)^2 = r^2 .

Since we are centered about the origin both and are zero. Thus the equation we have is:

x^2 + y^2 = 49 after simplifying 7^2

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Example Question #22 : How To Find The Equation Of A Circle

What is the equation of a circle with center (4, 2) and radius of r=3 ?

Possible Answers:

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

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

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

(x-4)^2+(y-2)^2 = 3

Correct answer:

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

Explanation:

The basic formula for a circle in the coordinate plane is (x-h)^2+(y-k)^2 = r^2 , where (h,k) is the center of the circle with radius .

Using this, we can simply substitute for , for , and for . Customarily, r^2 is simplified for the final equation.

----> (x-4)^2+(y-2)^2 = 9 .

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ACT Math : ACT Math

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #1 : How To Find The Equation Of A Circle

Example Question #1 : Circles

Example Question #2 : Circles

Example Question #811 : Act Math

Example Question #11 : How To Find The Equation Of A Circle

Example Question #12 : How To Find The Equation Of A Circle

Example Question #13 : How To Find The Equation Of A Circle

Example Question #21 : How To Find The Equation Of A Circle

Example Question #21 : How To Find The Equation Of A Circle

Example Question #22 : How To Find The Equation Of A Circle

All ACT Math Resources

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