SAT Math - 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.

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

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

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

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

None of the other choices gives a correct response.

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 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 , so . Substituting 0, 0, and 30 for , , and , respectively, the equation of the circle becomes

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

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

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.

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

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

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

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

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

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

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

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

Circle a

The above circle has area $64 \pi$ . Give its equation.

$(x- 8)^{2}+ (y - 8)^2 = 64$

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

$(x- 32)^{2}+ (y - 32)^2 = 32$

$(x- 32)^{2}+ (y - 32)^2 = 1,024$

$\textup{None of the other choices gives the correct response}$

Explanation

The equation of a circle on the coordinate plane is

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

where 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 $A = 64 \pi$ and solve for :

$\pi r^{2} = 64 \pi$

$\frac{ \pi r^{2} }{\pi} = \frac{64 \pi}{\pi}$

$r^{2} = 64$

$r = \sqrt{64} = 8$

As seen below, the horizontal and vertical distance from the origin to the center of the circle are both equal to this radius, and it is located in Quadrant I, so the center is :

Circle b

Setting , the equation of the circle becomes

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

$(x- 8)^{2}+ (y - 8)^2 = 64$

A circle has its origin at . The point is on the edge of the circle. What is the radius of the circle?

$r=\sqrt{74}$

$r=2\sqrt{18}$

$r=6\sqrt{2}$

$r=4\sqrt{21}$

There is not enough information to answer this question.

Explanation

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

$r=\sqrt{74}$

This radical cannot be reduced further.

Circle a

The provided figure shows a circle on the coordinate axes with its center at the origin. $\overarc {AB}$ is a $45^{\circ }$ arc with length $\frac{3}{4} \pi$

Give the equation of the circle.

$x^{2}+ y^2 =9$

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

$x^{2}+ y^2 =6$

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

$x^{2}+ y^2 =3$

Explanation

A $45^{\circ }$ arc of a circle represents $\frac{45}{360 } = \frac{1}{8}$ of the circle, so the length of the arc is one-eighth its circumference - or, equivalently, the circumference is the length $\frac{3}{4} \pi$ multiplied by 8. Therefore, the circumference is

$C = \frac{3}{4} \pi \cdot 8 = 6 \pi$ .

The equation of a circle on the coordinate plane is

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

where 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

$r = \frac{C}{2 \pi } = \frac{6 \pi }{2 \pi } = 3$

The center of the circle is , so . Substituting 0, 0, and 3 for , , and , respectively, the equation of the circle becomes

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

$x^{2}+ y^2 =9$ .

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?

$\sqrt{23}$

$\sqrt{21}$

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 = 52

_x_2 + _y_2 = 25

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

22 + _y_2 = 25

4 + _y_2 = 25

_y_2 = 21

y = ±√(21)

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

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

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

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

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

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

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

Explanation

The equation of the circle on the coordinate plane with radius and center 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 and 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 and 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 . 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 and $r ^{2} = 17$ in the circle equation:

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

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

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

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

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

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

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

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

Explanation

The equation of the circle on the coordinate plane with radius and center 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 and 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:

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

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

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

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

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

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

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

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

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

Explanation

The equation of the circle on the coordinate plane with radius and center 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 and , 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 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.

We have a square with length 2 sitting in the first quadrant with one corner touching the origin. If the square is inscribed inside a circle, find the equation of the circle.

Explanation

If the square is inscribed inside the circle, in means the center of the circle is at (1,1). We need to also find the radius of the circle, which happens to be the length from the corner of the square to it's center.