how do you find the domain and range of a circle?

Question

how do you find the domain and range of a circle.

Answer 1

Your domain and range of a circle depends of the radius of that circle. Since a circle is a set of points equidistant from a fixed point, your domain and range will have the same intervals. As proof, draw the center of the circle at the origin (0,0).  Then use a compass to draw a circle with (0,0) as your reference point. You will see the domain and range are the same.

 

 

Answer 2

The standard equation for a circle is (x-a)² + (y-b)² = r², where the center of the circle is at (a, b) and r is the radius. If you're given the equation of a circle that is not in standard form, complete the square for the x- and y-terms to get it in that form.

 

Then the domain will be a-r ≤ x ≤ a+r,

and the range will be b-r ≤ y ≤ b+r

 

The general equation for a circle is x² + y² + Ax + By + C = 0, where A, B, and C are constants.

 

To put this in standard form, you have to

 

1) gather x- and y-terms,

2) put the constant C on the other side,

3) complete the square in x and in y, and

4) factor those squares.

 

You'll then have the standard form, from which you can read off the domain and range as above. You can find details about this here: http://www.mathsisfun.com/algebra/circle-equations.html

 

1 and 2) x² + Ax + y² + By = -C

3) [x² + Ax + (A/2)²] + [y² + By + (B/2)²] = -C + (A² + B²)/4

 

If the constants on the right are zero or less, you're done, because the terms on the left will be squares and thus always positive. In these cases, you don't have a circle.

 

Assuming the constants add to a positive number, factor the x- and y- groups on the left to get

4) [x + (A/2)]² + [y + (B/2]² = (A² + B²)/4 - C

This is the equation for a circle whose center is at (-A/2, -B/2) and radius = square root of the mess on the right.

Notice that the squares on the left will always be positive.

Answer 3

All Reals