Posted by Jessica on .
The points (4, –5) and (– 4, 1) are endpoints of a diameter of a circle.
Find the center of the circle.

Math 
MathMate,
The midpoint Pm between two points P1(x1,y1) and P2(x2,y2) can be found by taking the mean of the x and ycoordinates.
Thus
xm=(x1+x2)/2 and
ym=(y1+y2)/2
and the midpoint (centre of the diameter)
Pm(xm,ym).
Can you take it from here and post the results for verification if you wish? 
Math 
Jessica,
I am going to try to. How can I graph this so I can see it?

Math 
MathMate,
You can graph the two end points, then calculate the midpoint according to the given formula. Plot the midpoint and draw a circle with a radius equal to the length between the midpoint and one of the two ends. If the circle passes through both endpoints of the diameter, and the midpoint falls on the diameter, your calculations are correct.

Math 
Jessica,
This is what I came up with, but I am still having a hard time graphing it.
Is this correct?
Midpoint formula =
(x1+x2)/2
(y1+y2/2
x = (44)/2 = 0
y = (5 +1)/2 = 2
(0,2) is the center?? 
Math 
MathMate,
Perfect!
Now give it a try to graph the points and the circle, if you could. 
Math 
Jessica,
I did it! The center is (0,2) on the graph after tracing the circle. Now i need help with the length of the radius of the circle. (Note that this is a distance.) Give the exact answer? What do I do here please? Thanks!

Math 
MathMate,
Here's a link to the graph that may help you do your own:
http://i263.photobucket.com/albums/ii157/mathmate/Jessica.jpg 
Math 
MathMate,
The radius would be the distance between the centre and one of the two endpoints.
The distance D between two known points P1(x1,y1) and P2(x2,y2) can be calculated using Pythagoras theorem:
D = sqrt((x2x1)^{2}+(y2y1)^{2})
Now put P1 as one of the endpoints of the diameter, and P2 as the centre of the circle that you have calculated, you should be able to find D.
The answer for the radius should be 5. 
Math 
Jessica,
Don't see how you get 5?
How do you square 4? 4*4 = 16 correct? So then how can R = 5? I don't get it...Thanks by the way for the graph! 
Math 
Jessica,
I think I see what you are now talkin about! is this correct?
D = sqrt ((x2x1)2+ (y2y1)2)
4 4 = 8
1 5 = 6
Half the distance between the two given points
d = diameter = sqrt (82 + 62) =sqrt (64+36) = sqrt (100 = 10
Radius = 5
NOW i need to know how to put this in standard form? If this is correct? 
Math 
MathMate,
put P1(0,2), and P2(4,5)
D = sqrt((x2x1)^{2}+(y2y1)^{2})
= sqrt((40)^{2}+(5  2)^{2})
= sqrt(4^{2}+(3)^{2})
= sqrt(16+9)
= sqrt(25)
= 5 
Math 
Jessica,
Would this be considered standard form?
(40)2 + (5 2)2) = r^2
(40)2 + (5 2)2) = 5^2 
Math 
MathMate,
Yes, this is the standard form for finding the distance between two given points.
The 2's after the parentheses are meant to be exponents, of course.
(40)^2 + (5 2)^2) = r^2
(40)^2 + (5 2)^2) = 5^2 
Math 
Jessica,
Thanks for all your help! You really explained this topic well! Thanks again, and have a terrific weekend!

Math 
MathMate,
Yes, what you have done is finding the distance between the end points, and halve the diameter to get the radius. This is OK (if I am teaching). The intention could be asking you to find the distance between the centre and one of the endpoints, as I have illustrated above.
In any case, understanding what you are doing is more important than just the answer. You have taken the time to understand, that is good! 
Math 
Jessica,
What do you mean by "OK if you are teaching? I was just wondering?

Math 
MathMate,
I would prefer to see the distance calculated directly between the centre and one of the endpoints, but I would most probably accept calculating the distance between endpoints of the diameter and halving it. The reason being that the question requires you to calculate the centre, so it is natural that the result be used to calculate the radius directly. So I said OK because it is the correct answer anyway.

Math 
Jessica,
Oh ok! Thanks again! You really know your stuff! Have a great weekend!

Math 
MathMate,
"OK if you are teaching" meant if I were the teacher.

Math 
MathMate,
It meant "if I were the teacher".

Math 
Godgift,
10N*30cm=15N*x
300Nm=15Nx
x=300cm/15N
x=20cm