Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Coordinate Geometry and Distance Formula

Date: 10/02/2002 at 16:50:01
From: Chelsey
Subject: Coordinate geometry using an equation and the distance 
formula

I need to find the center and radius of the following equations and 
then graph them:  

   (x-3)^2 + (y+4)^2 - 9 = 0

   x^2 + y^2 = 36

I tried to use the distance formula d= sqrt(x2-x1)^2 + (y1-y2)^2.  
I tried x-4 + y+16 = 9. I moved the -4 over and the 16 over and got 
-3. But I don't know if this is correct or if the -3 is the x or y 
coordinate. 

I used the distance formula theorem when I was given two sets of 
coordinates to find the radius of a circle, which I got right (that 
was easy). But my book doesn't explain how to find the center and 
radius, and graph it without being given any coordinates. I looked 
back in my old algebra I book but it doesn't explain anything about 
finding the center, radius, or coordinates.  

Thank you for helping me,
Chelsey


Date: 10/02/2002 at 17:07:35
From: Doctor Jerry
Subject: Re: Coordinate geometry using an equation and the distance 
formula

Hi Chelsey,

The first equation is

   (x-3)^2 + (y-(-4))^2 = 3^2.

You can think about the left side as the square of the distance from
(x,y) to (3,-4), which according to the equation is 3^2. So the
distance from (x,y) to (3,-4) is 3, and (x,y) is on a circle with
center (3,-4) and radius 3.

A circle with center at the point (h,k) and radius r is, using the
distance formula

   (x-h)^2 + (y-k)^2 = r^2

This is just like my first paragraph.

So,

   x^2 + y^2 = 36 = 6^2

is the equation of a circle with center (0,0) and radius 6.

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 10/05/2002 at 20:02:28
From: Chelsey
Subject: Coordinate geometry using an equation and the distance 
formula

The equation says -3, +4, why did you change it to -(-4)?  I know it 
is the same, but why did you do it that way? Is it because the 
distance formula is subtraction instead of addition? I don't see how 
you got coordinates of (0,0) for the second equation. Also, thanks 
for helping me.  

Chelsey


Date: 10/05/2002 at 20:58:00
From: Doctor Ian
Subject: Re: Coordinate geometry using an equation and the distance 
formula

Hi Chelsey,

Think about a circle with radius k centered at the origin. The 
equation of the circle is 

  x^2 + y^2 = k^2

To see why this is true, you can pick any point on the circle and drop 
a line from there to the x-axis. You'll form a right triangle whose 
sides will be x and y, and whose hypotenuse will be r. The Pythagorean 
theorem tells us that for any right triangle, with legs A and B, and 
hypotenuse C, it will be true that

  A^2 + B^2 = C^2

In the case of a circle, the hypotenuse is always the same - it's 
just the radius of the circle - and any (x,y) pair that you choose on 
the circle satisfies the Pythagorean theorem. 

Does that make sense? 

Now, if we move the circle n units to the right, the equation of the 
new circle will be 

  (x-n)^2 + y^2 = k^2

Why does the equation change this way? One way to think about it is 
this. Let's consider an arbitrary curve on the x-y plane:
  
       *             |
        *            |
            *        |
                   * |
                     |  *
    -----------------+-----*-------
                     |      
                     |
                     |       *
                     |          *

Suppose we move the curve to the right by one unit. This is the 
same as moving the y-axis to the left by one unit.

  
       *        .    |
        *       .    |
            *   .    |
                .  * |
                .    |  *  
    ------------.----+-----*-------
                .    |      
                .    |
                .    |       *
                .    |          *

Now, whatever used to be going on at (x,whatever) is going on at 
(x-1,whatever) instead. So using (x-1) for x everywhere is the same as 
shifting everything to the right by one unit. 

Does that make sense?

Similarly, using (x+1) for x is the same as shifting everything to the 
left. Using (y-1) for y shifts everything up, and using (y+1) for y 
shifts everything down. 

Now, what happens if we shift a circle to the right by one unit? The 
center, which used to be at (0,0), is now at (1,0). In general, if we 
shift the circle

  x^2 + y^2 = k^2

to the right by n units, and up by m units, we get the equation

  (x-n)^2 + (y-m)^2 = k^2

So we like this form, because it's practically a description of the 
circle: "The radius is k, and the center is at (n,m)."  

Does that answer your question? 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Conic Sections/Circles

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/