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SAT Math : SAT Mathematics

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Example Questions

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

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:

There is not enough information to answer this question.

$r=2\sqrt{18}$

$r=\sqrt{74}$

$r=4\sqrt{21}$

$r=6\sqrt{2}$

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 #1 : How To Find The Equation Of A Circle

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:

$40\pi$

$12\pi$

$80\pi$

$20\pi$

$160\pi$

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 #11 : How To Find The Equation Of A Circle

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

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

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

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

(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

Find the equation of the circle centered at (-5,8) with a radius of 3.

Possible Answers:

(x+5)^2+(y-8)^2=3

(x+5)^2+(y-8)^2=9

(x-5)^2+(y+8)^2=3

(x-5)^2+(y-8)^2=9

(x-5)^2+(y+8)^2=9

Correct answer:

(x+5)^2+(y-8)^2=9

Explanation:

Write the standard equation of a circle, where (a,b) is the center of the circle, and is the radius.

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

Substitute the point and radius.

(x+5)^2+(y-8)^2=9

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

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?

Possible Answers:

$\sqrt{23}$

$\sqrt{21}$

Correct answer:

$\sqrt{21}$

Explanation:

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

x² + y² = r²

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

x² + y² = 5²

x² + y² = 25

Now, the question asks for the positive y-coordinate when x = 2. To solve this, simply plug in for x:

2² + y² = 25

4 + y² = 25

y² = 21

y = ±√(21)

Since our answer will be positive, it must be √(21).

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Example Question #263 : New Sat

What is the equation of a circle with center (1,1) and a radius of 10?

Possible Answers:

(x-1)^2+(y-1)^2=100

(x-1)+(y-1)=100

x^2 + y^2 = 10

(x+1)^2+(y+1)^2=100

x^2 + y^2 = 100

Correct answer:

(x-1)^2+(y-1)^2=100

Explanation:

The general equation for a circle with center (h, k) and radius r is given by

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

In our case, our h-value is 1 and our k-value is 1. Our r-value is 10.

Substituting each of these values into the equation for a circle gives us

(x-1)^2+(y-1)^2=100

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Example Question #15 : Circles

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

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

Possible Answers:

$(2\pi, 1)$

$(\pi, 0)$

(1, 2)

None of the given answers.

$(0, \pi)$

Correct answer:

$(0, \pi)$

Explanation:

Remember that the general equation for a circle with center (h, k) and radius is (x-h)^2 + (y-k)^2 = r^2 .

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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Example Question #11 : Circles

A square on the coordinate plane has vertices at the points with coordinates (9,0) , (9,9) , (0, 9) , and (0,0) . Give the equation of the circle that circumscribes the square.

Possible Answers:

$\left ( x -9 \right )^{2} + \left ( y- 9\right )^{2} = 81$

$\left ( x - \frac{9}{2}\right )^{2} + \left ( y- \frac{9}{2}\right )^{2} =81$

$\left ( x - \frac{9}{2}\right )^{2} + \left ( y- \frac{9}{2}\right )^{2} = \frac{81}{2}$

$\left ( x -9 \right )^{2} + \left ( y- 9\right )^{2} = \frac{81}{2}$

$\left ( x - \frac{9}{2}\right )^{2} + \left ( y- \frac{9}{2}\right )^{2} = \frac{81}{4}$

Correct answer:

$\left ( x - \frac{9}{2}\right )^{2} + \left ( y- \frac{9}{2}\right )^{2} = \frac{81}{2}$

Explanation:

The equation of the circle on the coordinate plane with radius and center (h,k ) is

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

The figure referenced is below:

Circle x

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 (0,0) and (9,9) , 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 (0,0) 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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Example Question #12 : Circles

A square on the coordinate plane has vertices at the points with coordinates (7,0) , (-7, 0) , (0, 7) , and (0,-7) . Give the equation of the circle that circumscribes the square.

Possible Answers:

$x ^{2} +y ^{2} = 49$

$(x -7 )^{2} +(y - 7) ^{2} = \frac{49}{2}$

$(x -7 )^{2} +(y - 7) ^{2} = 49$

$x ^{2} +y ^{2} = \frac{49}{2}$

$x ^{2} +y ^{2} = 7$

Correct answer:

$x ^{2} +y ^{2} = 49$

Explanation:

The equation of the circle on the coordinate plane with radius and center (h,k ) is

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

The figure referenced is below:

Circle x

The center of the circle is the origin (0,0) ; the radius is 7. Therefore, setting h = k = 0 and r = 7 in the circle equation:

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

$(x-0)^{2} + (y-0)^{2} = 7 ^{2}$

$x ^{2} +y ^{2} = 49$

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

A square on the coordinate plane has vertices at the points with coordinates (0, 8) , (2, 0) , (10, 2) , and (8, 10 ) . Give the equation of the circle that circumscribes the square.

Possible Answers:

$(x-5)^{2} + (y-5)^{2} = 34$

$x^{2} + y^{2} = 34$

$x^{2} + y^{2} = 17$

$(x-5)^{2} + (y-5)^{2} = 68$

$(x-5)^{2} + (y-5)^{2} = 17$

Correct answer:

$(x-5)^{2} + (y-5)^{2} = 34$

Explanation:

The equation of the circle on the coordinate plane with radius and center (h,k ) is

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

The figure referenced is below:

Circle x

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 (2,0) and (8, 10 ) , we can set $x_{1} =2, y_{1} = 0 , x_{2} =8, y_{2} = 10$ in the midpoint formula:

$h =\frac{ x_{1}+ x_{2} }{2} = \frac{2+8}{2} = \frac{10}{2} = 5$

and

$k=\frac{ y_{1}+ y_{2} }{2} = \frac{0+ 10}{2} = \frac{10}{2} = 5$

The center of the circumscribed circle is therefore (5,5) .

The radius of the circle is the distance from this point to any of the vertices - we will use (2,0) . 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

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

Setting $x_{1} =2, y_{1} = 0, x_{2} =5, y_{2} = 5$ :

$r^{2}= (5-2)^{2} +(5-0)^{2}$

$=3^{2} +5^{2}$

=9+25

=34

Setting h = k = 5 and $r^{2} = 34$ in the circle equation:

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

$(x-5)^{2} + (y-5)^{2} = 34$

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SAT Math : SAT Mathematics

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

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

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

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

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

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

Example Question #263 : New Sat

Example Question #15 : Circles

Example Question #11 : Circles

Example Question #12 : Circles

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

All SAT Math Resources

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