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### 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

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
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)."

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

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