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SAT Math : Circles

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Example Question #231 : Coordinate Geometry

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

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

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

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 #232 : Coordinate Geometry

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=9

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

(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 #233 : Coordinate Geometry

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 #261 : New Sat

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

Possible Answers:

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

x^2 + y^2 = 10

x^2 + y^2 = 100

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

(x+1)^2+(y+1)^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 #235 : Coordinate Geometry

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:

$(2\pi, 1)$

$(\pi, 0)$

$(0, \pi)$

None of the given answers.

(1, 2)

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 #11 : 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:

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

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

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

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

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

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 #13 : Circles

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:

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

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

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

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

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

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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Example Question #14 : Circles

A square on the coordinate plane has as its vertices the points with coordinates (0,0) , (0,9) , (9,0) , and (9,9) . Give the equation of the circle inscribed inside this square.

Possible Answers:

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

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

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

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

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

Correct answer:

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

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:

Incircle 1

The center of the inscribed circle is the center of the square, which is where its diagonals intersect; this point is the common midpoint of the diagonals. The coordinates of the midpoint of the diagonal with endpoints at (0,0) and (9,9) can be found by setting $x_{1} = y_{1} = 0 , x_{2} = y_{2} = 9$ in the following midpoint formulas:

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

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

This point, $\left (\frac{9}{2}, \frac{9}{2} \right )$ , is the center of the circle. The radius can easily be seen to be half the length of one side; each side is 9 units long, so the radius is half this, or $\frac{9}{2}$ .

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

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

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

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

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Example Question #15 : Circles

A square on the coordinate plane has as its vertices the points with coordinates (0, 8) , (2, 0) , (10, 2) , and (8, 10) . Give the equation of the circle inscribed inside this square.

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 inscribed circle is the center of the square, which is where its diagonals intersect; this point is the common midpoint of the diagonals. The coordinates of the midpoint of the diagonal with endpoints at (2,0) and (8,10) can be found by setting $x_{1} = 2, y_{1} = 0 , x_{2} = 8, y_{2} = 10$ in the following midpoint formulas:

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

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

This point, $\left (5,5 \right )$ , is the center of the circle.

The inscribed circle passes through the midpoints of the four sides, so first, we locate one such midpoint. The midpoint of the side with endpoints at (2,0) and (0, 8) can be located setting $x_{1} = 2, y_{1} = 0 , x_{2} = 0, y_{2} = 8$ in the midpoint formulas:

$x_{M} = \frac{x_{1}+ x_{2}}{2} = \frac{2+0}{2} = \frac{2}{2} = 1$

$y_{M} = \frac{y_{1}+ y_{2}}{2} = \frac{0+8}{2} = \frac{8}{2} = 4$

One of the points on the circle is at (1, 4) . The radius is the distance from this point to the center at $\left (5,5 \right )$ ; since we only really need to find $r ^{2}$ , we can set $x_{1} =1 , y_{1} = 4 , x_{2} = y_{2} = 5$ in the following form of the distance formula:

$r ^{2} = (x_{2} - x_1) ^{2} + (y_{2} - y_1) ^{2}$

$r ^{2} = (5-1) ^{2} + (5-4) ^{2}$

$r ^{2} = 4 ^{2} +1 ^{2}$

$r ^{2} = 16+1$

$r ^{2} = 17$

Setting h = k= 5 and $r ^{2} = 17$ in the circle equation:

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

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

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SAT Math : Circles

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #231 : Coordinate Geometry

Example Question #232 : Coordinate Geometry

Example Question #233 : Coordinate Geometry

Example Question #261 : New Sat

Example Question #235 : Coordinate Geometry

Example Question #11 : Circles

Example Question #11 : Circles

Example Question #13 : Circles

Example Question #14 : Circles

Example Question #15 : Circles

All SAT Math Resources

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