Circles - ACT Math

The equation for a circle is (x - h)2 + (y - k)2 = r2, where (h,k) is the center of the circle and r is the radius.

We can eliminate (x+3) +(y+2) = 4 and (x+3)2 + (y+2)2 = 64. The first equation does not square the terms in parentheses, and the second refers to a center of (-3,-2) rather than (3,2).

Drawing the line y = -2 and a point at (3,2) for the center of the circle, we see that the only way the line could be tangent to the circle is if it touches the bottom-most part of the circle, directly under the center. The point of intersection will be (3,-2). From this, we can see that the distance from the center to this point of intersection is 4 units between (3,2) and (3,-2). This means the radius of the circle is 4.

Use the center point and the radius in the formula for a circle to find the final answer: (x - 3)2 + (y - 2)2 = 16

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

The distance between the endpoints of the diameter of the circle is:

To find the radius, we divide d (the length of the diameter) by two.

Then we substitute the value of r into the formula for the area of a circle.

The standard form equation of a circle is , where is the center of the circle and is equal to the radius. Thus, since we have the circle's standard form equation already given to us, we can ignore and , since all we need is .

The area of circle is equal to , which is equal to .

The formula for a circle is

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

The area of a circle:

Therefore:

The circle has a center at (5,5) and a radius of 2. Therefore, the equation is (x-5)2+(y-5)2=22, or (x-5)2+(y-5)2=4.

The equation of a circle is

The center is or, written another way . Substituting for and for , our formula becomes

Finally, the formula of the circle is

The formula for the equation of a circle is:

(x-h) 2 + (y-k)2 = r2

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 = 32

x2 + (y-4)2 = 9

The general equation of a circle is (x – h)2 + (y – k)2 = r2, 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 √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√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 = r2.

(x – (–2))2 + (y – 2)2 = (2√29)2 = 22(√29)2 = 4(29) = 116.

(x + 2)2 + (y – 2)2 = 116.

The answer is (x + 2)2 + (y – 2)2 = 116.

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.

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).

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

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

To find the y-coordinate, substitute into one of the equations. Let's use :

The center of our circle is therefore .

Now, recall that the general form for a circle with center at is

For our data, this means that our equation is:

We need to expand this equation to and then complete the square.

This brings us to .

We simplify this to .

Thus the radius is 7.

When finding the center and radius of circle , 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, 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 . 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 .

Center: Radius:

The general equation of a circle is .

According to the question, . Thus, .

The general equation for the area of a circle is.

When we plug in 13 for , we get our area to equal .

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

This radical cannot be reduced further.

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:

The general formula for a circle centered about points with a radius of is:

.

Since we are centered about the origin both and are zero. Thus the equation we have is:

after simplifying