The mid-point Pm between two points P1(x1,y1) and P2(x2,y2) can be found by taking the mean of the x- and y-coordinates.
and the mid-point (centre of the diameter)
Can you take it from here and post the results for verification if you wish?
I am going to try to. How can I graph this so I can see it?
You can graph the two end points, then calculate the mid-point according to the given formula. Plot the mid-point and draw a circle with a radius equal to the length between the mid-point and one of the two ends. If the circle passes through both end-points of the diameter, and the mid-point falls on the diameter, your calculations are correct.
This is what I came up with, but I am still having a hard time graphing it.
Is this correct?
Mid-point formula =
x = (4-4)/2 = 0
y = (-5 +1)/2 = -2
(0,-2) is the center??
Now give it a try to graph the points and the circle, if you could.
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!
Here's a link to the graph that may help you do your own:
The radius would be the distance between the centre and one of the two end-points.
The distance D between two known points P1(x1,y1) and P2(x2,y2) can be calculated using Pythagoras theorem:
D = sqrt((x2-x1)2+(y2-y1)2)
Now put P1 as one of the end-points 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.
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!
I think I see what you are now talkin about! is this correct?
D = sqrt ((x2-x1)2+ (y2-y1)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?
put P1(0,-2), and P2(4,-5)
D = sqrt((x2-x1)2+(y2-y1)2)
= sqrt((4-0)2+(-5 - -2)2)
Would this be considered standard form?
(4-0)2 + (-5- -2)2) = r^2
(4-0)2 + (-5- -2)2) = 5^2
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.
(4-0)^2 + (-5- -2)^2) = r^2
(4-0)^2 + (-5- -2)^2) = 5^2
Thanks for all your help! You really explained this topic well! Thanks again, and have a terrific weekend!
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 end-points, 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!
What do you mean by "OK if you are teaching? I was just wondering?
I would prefer to see the distance calculated directly between the centre and one of the end-points, but I would most probably accept calculating the distance between end-points 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.
Oh ok! Thanks again! You really know your stuff! Have a great weekend!
"OK if you are teaching" meant if I were the teacher.
It meant "if I were the teacher".
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