Arcs and Intercept - PSAT Math

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 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 $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 left ( x-4 right $)^{2}$+left ( y-3 right $)^{2}$=49.

Thus the radius is 7.

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:

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

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 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 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.