ACT Math - How to find the equation of a circle

Find the equation of the circle with center coordinates of and a radius of .

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

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

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

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

Explanation

The equation of a circle is $(x-h)^{^{2}} + (y-k)^{^{2}}=r^{^{2}}$

The center is or, written another way . Substituting for and for , our formula becomes $(x-2)^{2}+(y-4)^{2}=3^{2}$

Finally, the formula of the circle is $(x-2)^{2}+(y-4)^{2}=9$

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).

The endpoints of a diameter of circle A are located at points and . What is the area of the circle?

$20\pi$

$40\pi$

$80\pi$

$160\pi$

$12\pi$

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 Actmath_7_113_q1

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.

$A=\pi r^{2}=\pi ({2\sqrt{5}})^{2}=20\pi$

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.

A circle exists entirely in the first quadrant such that it intersects the -axis at . If the circle intersects the -axis in at least one point, what is the area of the circle?

$36\pi$

$9\pi$

$144\pi$

$6\pi$

$Not\ enough\ information$

Explanation

We are given two very important pieces of information. The first is that the circle exists entirely in the first quadrant, the second is that it intersects both the - and -axis.

The fact that it is entirely in the first quadrant means that it cannot go past the two axes. For a circle to intersect the -axis in more than one point, it would necessarily move into another quadrant. Therefore, we can conclude it intersects in exactly one point.

The intersection of the circle with must also be tangential, since it can only intersect in one point. We can thus conclude that the circle must have both - and - intercepts equal to 6 and have a center of .

This leaves us with a radius of 6 and an area of:

$\pi *radius^{2}=\pi*6^{2} = 36\pi$

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.

Now use the equation of the circle with the center and .

We get

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

y_1 = 4_x + 3

and

y_2 = 5_x + 44?

None of the other answers

(x - 3)2 + (y - 44)2 = 144

(x - 22)2 + (y - 3)2 = 12

(x - 41)2 + (y - 161)2 = 144

(x + 41)2 + (y + 161)2 = 144

Explanation

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

4_x_ + 3 = 5_x_ + 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 = 122 or (x + 41)2 + (y + 161)2 = 144

In the standard coordinate plane, what is the radius and the center of the circle $(x-5)^{2}+(y+4)^{2}=27$ ?

$Center:(5,-4)\ Radius: 3\sqrt{3}$

$Center:(-4,5)\ Radius: 3\sqrt{3}$

$Center:(5,-4)\ Radius: 27$

$Center:(-5,4)\ Radius:27$

$Center:(-5,4)\ Radius: 3\sqrt{3}$

Explanation

When finding the center and radius of circle $(x-h)^{2}+(y-k)^{2}=r^{2}$ , the center is and the radius is . Notice that they are not negative even though in the equation they have negative signs in front. This becomes important when dealing with real numbers. Also, notice the square of .

Our circle, $(x-5)^{2}+(y+4)^{2}=27$ has the same principles applied as the above principle, therefore is our center. Notice how the numbers signs have been switched. This is the case for all circles due to the negative in the base equation above.

To find the radius of a circle, you must take the number the equation is equal to and square root it. This is due to the square of mentioned above. The $\sqrt{27}=3\sqrt{3}$ . Use the least common multiples of 27 to find that three 3’s make up 27. Take two threes out as the square root of a number multiplied by itself is itself. This leaves one 3 under the radical. Therefore our radius is $3\sqrt{3}$ .

Center: Radius: $3\sqrt{3}$

A circle is centered on point . The area of the circle is $9\pi$ . What is the equation of the circle?

Explanation

The formula for a circle is

is the coordinate of the center of the circle, therefore and .

The area of a circle:

$r^2\pi = 9\pi$

Therefore:

Which of the following equations describes all the points (x, y) in a coordinate plane that are five units away from the point (–3, 6)?

(x – 3)2 + (y + 6)2 = 5

(x + 3)2 + (y – 6)2 = 25

(x – 3)2 + (y + 6)2 = 25

y + 6 = 5 – (x – 3)2

(x – 3)2 – (y + 6)2 = 25

Explanation

We are trying to find an equation for all of the points that are the same distance (5 units) from (–3, 6). The locus of all points equidistant from a single point is a circle. In other words, we need to find an equation of a circle. The center of the circle will be (–3, 6), and the radius, which is the distance from (–3,6), will be 5.

The standard form of a circle is given below:

(x – h)2 + (y – k)2 = r2, where the center is located at (h, k) and r is the length of the radius.

In this case, h will be –3, k will be 6, and r will be 5.

(x – (–3))2 + (y – 6)2 = 52

(x + 3)2 + (y – 6)2 = 25

The answer is (x + 3)2 + (y – 6)2 = 25.

How to find the equation of a circle

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