Question

Write an equation of a circle with the given diameter with points A and B

Answer 1

The general formula for a circle with radius \(r\) and center \((h,k)\) is \[(x-h)^2+(y-k)^2=r^2\]

Answer 2

Remember that \[d = 2r\]

Answer 3

Find the radius of your circle from that, and then plug in the coordinates of the center. That's all there is to it.

Answer 4

im completely lost on where to even begin with this

Answer 5

The problem tells you the diameter of the desired circle, right? The diameter is related to the radius by the equation \[d = 2r\]Plug in your diameter and solve for the radius. You can do that, right?

Answer 6

Sorry, I didn't read the blurry screen capture closely enough. You need to find the distance between those two points first to get the diameter!

So, you have two points which are the endpoints of a diameter: call the points \((x_1,y_1) \text{ and }(x_2,y_2)\)

The distance between them (which is a diameter of the circle) is given by
\[d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\]

If this looks suspiciously like the Pythagorean theorem, that's because it is!

Now that you have the diameter in hand, divide it by 2 to find the radius. Now you can plug in the numbers in the original formula :-)

Answer 7

oh ok that is what threw me off

Answer 8

I'll do an example for you.

Circle with center (1,2) has diameter going from (1,0) to (1,4)

\[d = \sqrt{(1-1)^2 + (4-0)^2} = \sqrt{4^2} = 4\]\[d = 2r\]\[r = \frac{d}2 = \frac{4}{2} = 2\]Circle is \[(x-1)^2+(y-2)^2=2^2\]\[(x-1)^2+(y-2)^2=4\]

Answer 9

\[d=\sqrt{(8-0)^2+(6-0)^2}\]
\[\sqrt{64+36}\]
\[\sqrt{100}=10\]

Answer 10

good so far...

Answer 11

d=2r
10=2r
5=r

Answer 12

you're on a roll...

Answer 13

Sigh. I let you down again! Now you need to find the midpoint of the diameter to use as your center.

Midpoint of a line segment has coordinates which are just the average of the x coordinates and the average of the y coordinates.

midpoint of (0,0), (6,2) = (3,1)

Answer 14

Too much blood in my caffeine stream this morning :-)

Answer 15

That would be right except that I didn't use the points from your diameter in my example...

Your diameter endpoints are (0,0) and (8,6), so your midpoint aka the center is at...

Answer 16

@whpalmer4 where did the (3,1) come from

Answer 17

I'm making up different examples so as not to actually do the problem for you...

Answer 18

ohhh ok so \[(x-h)^2+(y-k)^2=r^2\]
\[(x-h)^2+(y-k)^2=5^2\]
center is (4,3)

Answer 19

right, but you need to plug in the correct values of \(h\), \(k\) to make that equation have its center at \((4,3)\)...

Answer 20

\[(x-4)^2+(y-3)^2=25\]

Answer 21

Yes, that is correct! Let's test it, just to be sure:

\[(8-4)^2+(6-3)^2 = 25\]\[4^2+3^2=25\]\[16+9=25\checkmark\]\[(0-4)^2+(0-3)^2=25\]\[(-4)^2+(-3)^2=25\]\[16+9=25\checkmark\]So it has a center in the right place and goes through the endpoints of the diameter. That's what we want!

Answer 22

hooray :) thanks @whpalmer4

Answer 23

I'll check your answer for the other one if you post it.

Answer 24

ok ill work it out and post

Answer 25

\[d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y _{1})^2}\]
\[d=\sqrt{(5-1)^2+(5-1)^2}\]
\[d=\sqrt{4^2+4^2}\]
\[d=\sqrt{16+16}\]
\[d=\sqrt{36}\]
d=6
\[6=2r\]
r=3

Answer 26

16+16=?

Answer 27

yes, the problem would be neater if it turned out to equal 36, but it doesn't :-)

Answer 28

dang lol
my adding is bad

Answer 29

\[d=\sqrt{32}\]

Answer 30

Well, I'll continue the problem with that mistake, just to show you how my checking would catch it at the end...

\[(x-3)^2+(y-3)^2 = 3^2\]Plug in one of our diameter endpoints:
\[(5-3)^2+(5-3)^2 = 3^2\]\[4+4 = 9\]Not quite. Must have done something wrong!

Answer 31

you can simplify \(\sqrt{32} = \sqrt{16*2} = \sqrt{4*4*2} = \)

Answer 32

'yea i noticed that simplifying radicals is difficult for me

Answer 33

im not sure what to do with that

Answer 34

\[\sqrt{a*a} = a\]so long as \(a>0\)

Answer 35

so we have \[\sqrt{4*4*2} = \sqrt{4*4}*\sqrt{2} = 4*\sqrt{2}=4\sqrt{2}\]

Answer 36

It's not essential for this problem — you could use \[r=\frac{\sqrt{32}}2\] and then you'd get \[r^2 = \frac{(\sqrt{32})^2}{2^2}\]can you simplify that to a number?

Answer 37

ok so \[r^2=\frac{ 32 }{ 4 }\]

Answer 38

kinda on the right track?

Answer 39

Exactly on the right track, just haven't quite pulled into the station :-)

Answer 40

\[r^2=8\]

Answer 41

You might watch this video for some help on simplifying square roots:
https://www.khanacademy.org/math/algebra/exponent-equations/simplifying-radical-expressions/v/simplifying-square-roots

Yes, \(r^2=8\) is correct. And the rest of the equation is...

Answer 42

(x-h)^2+(y-k)^2=8

Answer 43

but with the values of h, k please...

Answer 44

There are an infinite number of sets of \(h,k\) that make a circle with radius 8. I want the unique set that makes a circle with radius 8 and a center at the midpoint of the line segment from (1,1) to (5,5)

Answer 45

(5-2)^2+(1-2)^2=8
3^2-1^2=8
9-1=8

Answer 46

i think that is right

Answer 47

No, sadly, it is not.
\[(5-2)^2+(1-2)^2=8\]\[3^2-1^2=8\]
You didn't simplify that right. \[(5-2)^2+(1-2)^2 = (3)^2 + (-1)^2 = 9+1 = 10\]

Answer 48

What's the midpoint of the line segment from (1,1) to (5,5)?

Answer 49

(show your work, I have a suspicion about what you are doing wrong)

Answer 50

(it wouldn't matter if one endpoint was (0,0) as in the previous problem)

Answer 51

here's a graphical hint:

Answer 52

\[(5-3)^2+(5-3)^2=8\]\[(2^2)+(2^2)=8\]
\[4+4=8\]

sorry i forgot how to find average lol

Answer 53

i remember add the numbers then divide by # of terms

Answer 54

Yeah, that's what I thought — you did (5-1)/2 = 2, right? It doesn't matter with one endpoint being 0, of course, as 5-0 = 5+0

Answer 55

that is what i did lol -_-

Answer 56

it's funny how many kids I help who are unwilling to show their work when they are getting a wrong answer. they'll describe their work, and often they have the right concept of how to get to the answer, but the issue is that they are making a mistake in applying the concept, and the exact identity of the mistake is almost never apparent from listening to their description of the work they did :-)

Answer 57

i cant believe i forgot how to find average lol thats terrible

Answer 58

Well, actually, your mistake isn't so unreasonable. It is as far as finding the average goes, I can't deny that :-) but as a way to find the midpoint, it actually works, just didn't do the final step!

say we have our two points (1,1) and (5,5). You took the difference of the x coordinates, divided by 2, and stopped. If you had merely adding that half-difference to the x coordinate of the left-hand point, and done the same with the y coordinates, you would have gotten exactly the right answer.

Answer 59

in other words, half the distance between the points, added to the "starting" point, takes you to the midpoint of the journey.

Answer 60

think of it as a road trip from point A to B to C. If you want to figure out the halfway point between B and C, taking the mileage between them and dividing it by 2 gives you the wrong answer if you don't do anything more, but if you mark off that distance down the road between B and C, you get the right answer.

Answer 61

all that extra kinda confused me ^^

Answer 62

Look at the diagram I sent. Distance between the two points (measured along the x axis) is 4: 5-1=4. Distance between the two points (measured along the y axis) is also 4: 5-1 =4.

Half of that distance between the x coordinates is 4/2 = 2. If we add 2 to the starting x-coordinate (the one for the point on the left, that gives us 1+2 = 3. If we repeat the same procedure for the y coordinates, we also have 1+2 =3, giving us a midpoint of (1+2,1+2) = (3,3).


We could also show it algebraically. If our two x coordinates are \(x_1,x_2\), then the average is
\[\frac{x_1+x_2}{2}\]But that's equivalent to
\[\frac{x_1+x_2}{2} = \frac{x_2-x_1}2 + x_1\]Multiply both sides by 2\[x_1+x_2 = x_2-x_1 + 2x_1\]Collect like terms on each side\[x_1+x_2 = x_2+x_1\]

What you computed was \[\frac{x_2-x_1}{2}\]You just didn't add the \(+x_1\) to that.

But, if it's confusing you, forget I mentioned it, and go back and flog yourself with a wet noodle for forgetting how to do the average of two numbers :-)

Answer 63

ooh ok i think i kinda understand a little