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SAT Mathematics : Applying the Equation of a Circle

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Example Question #1 : Applying The Equation Of A Circle

What is the equation for a circle of radius 12, centered at the intersection of the two lines:

$y_{1}=4x+3$

and

$y_{2}=5x+44$ ?

Possible Answers:

$(x+41)^{2}+(y+161)^{2}=141$

$(x-22)^{2}+(y-3)^{2}=12$

$(x-41)^{2}+(y-161)^{2}=144$

None of the other answers

Correct answer:

$(x+41)^{2}+(y+161)^{2}=141$

Explanation:

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

$4x+3=5x+44;\, 3=x+44;\, -41=x$

To find the y-coordinate, substitute into one of the equations. Let's use $y_{1}$ :

y=4*-41+3=-164+3=-161

The center of our circle is therefore: (–41, –161).

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:

$(x+41)^{2}+(y+161)^{2}=12^{2}$ or $(x+41)^{2}+(y+161)^{2}=144$

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Example Question #11 : Circles

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 $x^{2}-8x+y^{2}-6y=24$ and then complete the square.

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

We simplify this to $(x-4)^{2}+(y-3)^{2}=49$ .

Thus the radius is 7.

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Example Question #12 : 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:

$r=6\sqrt{2}$

$r=2\sqrt{18}$

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 #13 : Circles

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:

$\sqrt{(21)}$

$\sqrt{(23)}$

Correct answer:

$\sqrt{(21)}$

Explanation:

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

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

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

$x^{2}+y^{2}=5^{2}$

$x^{2}+y^{2}=25$

Now, the question asks for the positive y-coordinate when . To solve this, simply plugin for :

$2^{2}+y^{2}=25$

$4+y^{2}=25$

$y^{2}=21$

$y=\pm \sqrt{(21)}$

Since our answer will be positive, it must be $\sqrt{(21)}$ .

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Example Question #14 : 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:

$(\pi, 0)$

$(0, \pi)$

$(2\pi, 1)$

(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 #6 : Applying The Equation Of A Circle

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:

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

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

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

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

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:

Screen shot 2020 09 29 at 11.32.47 am

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 #15 : Circles

Screen shot 2020 09 29 at 11.34.42 am

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:

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

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

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

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

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 #16 : Circles

Screen shot 2020 09 29 at 11.36.22 am

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:

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

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

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

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

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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Example Question #17 : Circles

If the center of a circle is at (0,4) and the diameter of the circle is 6, what is the equation of that circle?

Possible Answers:

$(x-4)^{2+y^{2}}=9$

$(x-4)^{2}+y^{2}=36$

$x^{2}+(y-4)^{2}=9$

$x^{2}+(y-4)^{2}=36$

Correct answer:

$x^{2}+(y-4)^{2}=9$

Explanation:

The formula for the equation of a circle is:

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

Where (h,k) is the center of the circle.

h=0 and k=4

and diameter = 6 therefore radius = 3

$(x-0)^{2}+(y-4)^{2}=3^{2}$

$x^{2}+(y-4)^{2}=9$

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Example Question #8 : Applying The Equation Of A Circle

Circle A is given by the equation $(x-4)^{2}+(y+3)^{2}=29$ . Circle A is shifted up five units and left by six units. Then, its radius is doubled. What is the new equation for circle A?

Possible Answers:

$(x-2)^{2}+(y+2)^{2}=58$

$(x-10)^{2}+(y+8)^{2}=116$

$(x+2)^{2}+(y-2)^{2}=58$

$(x+2)^{2}+(y-2)^{2}=116$

Correct answer:

$(x+2)^{2}+(y-2)^{2}=116$

Explanation:

The general equation of a circle is $(x-h)^{2}+(y-k)^{2}=r^{2}$ , where (h, k) represents the location of the circle's center, and r represents the length of its radius.

Circle A first has the equation of $(x-4)^{2}+(y+3)^{2}=29$ . This means that its center must be located at (4, –3), and its radius is $\sqrt{29}$ .

We are then told that circle A is shifted up five units and then left by six units. This means that the y-coordinate of the center would increase by five, and the x-coordinate of the center would decrease by 6. Thus, the new center would be located at (4 – 6, –3 + 5), or (–2, 2).

We are then told that the radius of circle A is doubled, which means its new radius is $2\sqrt{29}$ .

Now, that we have circle A's new center and radius, we can write its general equation using $(x-h)^{2}+(y-k)^{2}=r^{2}$ .

$(x-(-2))^{2}+(y-2)^{2}=(2\sqrt{29})^{2}=2^{2}(\sqrt{29})^{2}=4(29)=116$ .

$(x+2)^{2}+(y-2)^{2}=116$ .

The answer is $(x+2)^{2}+(y-2)^{2}=116$ .

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SAT Mathematics : Applying the Equation of a Circle

Study concepts, example questions & explanations for SAT Mathematics

All SAT Mathematics Resources

Example Questions

Example Question #1 : Applying The Equation Of A Circle

Example Question #11 : Circles

Example Question #12 : Circles

Example Question #13 : Circles

Example Question #14 : Circles

Example Question #6 : Applying The Equation Of A Circle

Example Question #15 : Circles

Example Question #16 : Circles

Example Question #17 : Circles

Example Question #8 : Applying The Equation Of A Circle

All SAT Mathematics Resources

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