# quation of a circle

• Mar 16th 2008, 01:00 PM
jwein4492
quation of a circle
The equation of circle A is (x-4)2+(y-2)2=16 and the equation of circle B is (x+6)2+(y+3)2=9. what is the distance between the centers of these circles?
• Mar 16th 2008, 01:07 PM
galactus
One is centered at (4,2) and the other at (-6,-3).

Use the distance formula and see.
• Mar 16th 2008, 01:07 PM
topsquark
Quote:

Originally Posted by jwein4492
The equation of circle A is (x-4)2+(y-2)2=16 and the equation of circle B is (x+6)2+(y+3)2=9. what is the distance between the centers of these circles?

The general form for a circle is
$(x - h)^2 + (y - k)^2 = r^2$
where the center is (h, k). What can you do with this?

-Dan
• Mar 16th 2008, 01:16 PM
jwein4492
thanks now I have one that is
the equation of circle P is (x-3)2+(y+2)2=34 and equation of circle Q is (x-4)2+(y+5)=52 What is the distance between the centers of these circles?

And Use the distance formula to find the distance between the 2 centers. But I dont understand how.

its the square root of (x2-x) to the second + (y2-y)to the second. HELP!!!
• Mar 16th 2008, 01:23 PM
topsquark
Quote:

Originally Posted by jwein4492
The equation of circle A is (x-4)2+(y-2)2=16 and the equation of circle B is (x+6)2+(y+3)2=9. what is the distance between the centers of these circles?

Quote:

Originally Posted by topsquark
The general form for a circle is
$(x - h)^2 + (y - k)^2 = r^2$
where the center is (h, k). What can you do with this?

-Dan

The first circle has a center at (4, 2). The second circle has a center at (-6, -3).

The distance between any two points $(x_1, y_1)$ and $(x_2, y_2)$ is
$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Can you take it from here?

-Dan
• Mar 16th 2008, 01:31 PM
jwein4492
No
I dont understandn the square root stuff.
• Mar 16th 2008, 01:34 PM
topsquark
Quote:

Originally Posted by jwein4492
I dont understandn the square root stuff.

$\sqrt{(-6 - 4)^2 + (-3 - 2)^2}$

It's just a plug'n'chug problem from here. Are you having problems simplifying the answer?

-Dan
• Mar 16th 2008, 01:37 PM
jwein4492
yes
yes on simplifying the answer and showing my work on it?
my teacher is really hard to understand and doesnt explain very well.
• Mar 16th 2008, 01:40 PM
topsquark
Quote:

Originally Posted by jwein4492
yes on simplifying the answer and showing my work on it?
my teacher is really hard to understand and doesnt explain very well.

$\sqrt{125} = \sqrt{25 \cdot 5} = \sqrt{25} \cdot \sqrt{5} = 5 \sqrt{5}$

-Dan
• Mar 16th 2008, 01:44 PM
jwein4492
cool
thanks so much i am sure we will talk again.