To find the  intercepts,  is set equal to . This yields:

And finally

It is important to realize that both  and  must be included because  is also equal to . Finally, these are put into their point forms,  and .

To find the x-intercepts of an equation, you can just set the y value of the equation equal to zero.  Thus you get, for our data:

Now, you can divide everything by  to simplify your equation:

Luckily, this is an easy equation to factor:

Based on this, you know that the two intercepts must be where  and .  Thus, the largest intercept is .

To find the x-intercepts of an equation, you can set its y value equal to zero.  Thus, you get for our equation:

Now, factor out all common factors:

From this, you can further factor:

Thus, the x-intercepts of our equation are ,, and .  The sum of these values is .

The y-intercept is the constant at the end of the equation. Thus for our equation the y-intercept is 7

The x-intercept is the value of the linear equation with y = 0 (this means the line will be on the x-axis when y is zero).

Thus we plug 0 in for y and solve for x.

.

Now put it in an ordered pair, remember y = 0:

To find the  and  intercept of a linear equation, find the points where  and  are equal to zero.

To do this, plug in zero for either variable and then solve for the other.

this yields: 

To find the  and  intercepts of an equation, set each variable to zero (one at a time) and solve for the other variable.

Next, set   to zero:

Now put these two sets of points into two ordered pairs:

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

If a circle is tangent to the x-axis at (3,0), this means it touches the x-axis at that point. If a circle is tangent to the y-axis at (0,3), this means it touches the y-axis at that point. Given these two points, we can determine the center and the radius of the circle. The center of the circle must be equidistant from any of the points on the circumference. This means that both (0,3) and (3,0) are the same distance from the center. If we draw these points on a coodinate plane, it becomes apparent that the center of the circle must be (3,3). This point is exactly three units from each of the given points, indicating that the radius of the circle is 3.

When we input this information into the formula for a circle, we get (x – 3)² + (y – 3)² = 9.