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SAT Math : Coordinate Geometry

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Example Questions

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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Example Question #13 : 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:

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

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

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

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

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

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 #14 : 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:

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

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

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

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

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

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

Circle a

The above figure shows a circle on the coordinate axes with its center at the origin. $\overarc {ACB}$ has length $35\pi$ .

Give the equation of the circle.

Possible Answers:

$x^{2}+ y^2 =30$

$x^{2}+ y^2 =900$

None of the other choices gives a correct response.

$x^{2}+ y^2 =1,764$

$x^{2}+ y^2 =42$

Correct answer:

$x^{2}+ y^2 =900$

Explanation:

$\overarc {A B}$ has measure $150 ^{\circ }$ , so $\overarc {ACB}$ , its corresponding major arc, measures $(360 - 150) ^{\circ } = 210 ^{\circ }$ , making it $\frac{210}{360 } = \frac{7}{12}$ of the circle. The length of $\overarc {ACB}$ , $21 \pi$ , is seven-twelfths its circumference, so set up the equation and solve for :

$\frac{7}{12} C = 35 \pi$

$\frac{12} {7} \cdot \frac{7}{12} C = \frac{12} {7} \cdot 35 \pi$

$C = 60\pi$

The equation of a circle on the coordinate plane is

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

where (h, k) 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

$r = \frac{C}{2 \pi } = \frac{60 \pi }{2 \pi } = 30$

The center of the circle is (0,0) , so h = k = 0 . Substituting 0, 0, and 30 for , , and , respectively, the equation of the circle becomes

$(x- 0)^{2}+ (y - 0)^2 =30^{2}$ ,

$x^{2}+ y^2 =900$ .

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Example Question #241 : Coordinate Geometry

Circle a

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.

Possible Answers:

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

$x^{2}+ y^2 =36$

$x^{2}+ y^2 =16$

$x^{2}+ y^2 =72$

$x^{2}+ y^2 =4$

Correct answer:

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

Explanation:

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 (h, k) 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 (0,0) , so h = k = 0 . Substituting 0, 0, and 8 for , , and , respectively, the equation of the circle becomes

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

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

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

Circle b

The above circle has area $36 \pi$ . Give its equation.

Possible Answers:

$x ^{2}+ (y - 6)^2 = 36$

(x - 16)^2+ y^2= 36

$x ^{2}+ (y - 16)^2 = 256$

(x - 16)^2+ y^2= 256

None of the other choices gives the correct response

Correct answer:

$x ^{2}+ (y - 6)^2 = 36$

Explanation:

The equation of a circle on the coordinate plane is

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

where (h, k) are the coordinates of the center and is the radius.

The area and the radius of a circle are related by the formula

$\pi r^{2} = A$

Set $A = 36\pi$ and solve for :

$\pi r^{2} = 36\pi$

$\frac{ \pi r^{2} }{\pi} = \frac{36\pi}{\pi}$

$r^{2} = 36$

$r = \sqrt{36} = 6$ .

The center of the circle lies on the -axis, so h = 0 . Also, the center is 6 units above the origin, so k = 6 . Setting h=0 ,k=6,r=6 , the equation becomes

$(x- 0)^{2}+ (y - 6)^2 = 6^{2}$

$x ^{2}+ (y - 6)^2 = 36$ .

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Example Question #22 : Circles

Circle b

The above circle has area $100 \pi$ . Give its equation.

Possible Answers:

None of the other choices gives the correct response

$(x-10)^{2}+y^2 = 100$

$(x+10)^{2}+y^2 = 100$

$(x-50)^{2}+y^2 = 2,500$

$(x+50)^{2}+y^2 = 2,500$

Correct answer:

$(x+10)^{2}+y^2 = 100$

Explanation:

The equation of a circle on the coordinate plane is

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

where (h, k) are the coordinates of the center and is the radius.

The area and the radius of a circle are related by the formula

$\pi r^{2} = A$

Set $A = 100\pi$ and solve for :

$\pi r^{2} = 100\pi$

$\frac{ \pi r^{2} }{\pi} = \frac{100\pi}{\pi}$

$r^{2} = 100$

$r = \sqrt{100} = 10$ .

The center of the circle lies on the -axis, so k = 0 . Also, the center is 10 unites left of the origin, so h = -10 . Setting h= -10,k=0,r =10 accordingly, the equation becomes

$(x- (-10 ))^{2}+ (y - 0)^2 = 10^{2}$

$(x+10)^{2}+y^2 = 100$ .

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Example Question #23 : Circles

Circle a

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.

Possible Answers:

$x^{2}+ y^2 =28$

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

$x^{2}+ y^2 =112$

$x^{2}+ y^2 =32$

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

Correct answer:

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

Explanation:

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 (h, k) 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 (0,0) , so h = k = 0 . Substituting 0, 0, and 56 for , , and $r^{2}$ , respectively, the equation of the circle becomes

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

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

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SAT Math : Coordinate Geometry

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #11 : Circles

Example Question #12 : Circles

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

Example Question #13 : Circles

Example Question #14 : Circles

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

Example Question #241 : Coordinate Geometry

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

Example Question #22 : Circles

Example Question #23 : Circles

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