Just for fun - circles

[Acidic]

Lets say there are the points A(2,2) and B(3,4) and C(-1,2).
A circle just touches them all. Find the centre of the circle and it's radius.

I've just made this number up, so I don't know of the circle exists.

[wossname]

The circle definitely does exist according to my logic: Draw a circle that passes through two of the points and then make the circle larger until it intersects the third (making sure the circle never leaves the first two points. You might have to flip the circle over sometimes to make sure the 3rd point is on the same side.

The only time the circle is impossible is when the 3 points are on a line... The circle would have to be infinitely large.

A(0,1), B(0,2), C(0,3).

Can't think of a formula though?

[NoteMe]

Yeah as long as they are not in line you can find a circle. You just have to find out where the centre is. The only way I can think of right now is to find a point that has the same lenght from all points.

[NoteMe]

I have been drawing a lot of circles now. And I always make it a true circle with only 3 points. Two of the points will always be in a line. And then it isn't hard ot make the 3 dot connect...

[Acidic]

NoteMe saying:
The only way I can think of right now is to find a point that has the same lenght from all points.

triggered something off
firstly I simplified the question. I made the point:
A = (0,1)
B = (1,0)
C = (-1,0)

Now it is obvious what the circle is but it might still help us with more complex questions.

I drew the line y=x
that line gives all the points where A&B are equi-distant. Then all I had to do was find where on that line the distance to all three points are the same. Hmm. maybe I should put on my Maths socks.

I'll have to ponder this a bit longer.

http://acid.freewebpage.org/circle.jpg

[Acidic]

Yess, got it sorted

You draw all the lines showing the equidistant points between A&B, A&C and B&C (though only two lines are really needed)

Where these lines intersect, thats where the centre is. The radius is then quite easy to find.

[NoteMe]

I am not 100% sure what you are talking about. But I was thinking about a thing. If you draw the lines from:

A->B
B->C
C->A

And you draw the normals between them. Will they then intersect in the centre?? Or was that what you just said???

[Acidic]

basically yes:

here what I wrote about ten minutes ago but have had a whole world of trouble to post:

got it
right:
draw the points
A = (2,2)
B = (3,4)
C = (-1,2)

the line showing equidistant points between B & C is a vertical line at x=0.5, draw that.

draw a straight line connecting A & B (y=2x-2).

find the middle point (2.5,3)

find the gradient of the normal (-0.5)

then the eqn for the normal is y=.5x + C
we know that it passes the point (2.5,3)
so:
3=-.5*2.5 + C
C = 4.125
so the equation for the normal is:
y= -.5c + 4.125

to see where it intersects the other line put x=0.5
y=-0.5*.5+4.125
y=3.875
therefore the centre is at (0.5,3.875)

now the radius, using pythagoras theorem, the distance from point C to (0.5,3.785) is:
sqrt[ (0.5+1)2 + (3.875-2)2 ]
= 2.031

so centre at (0.5,3.875)
radius: 2.031

[Acidic]

Here's the problems I had in posting the above post.

I solved in on papar. While writing it on VBF, my mum calls my mobile asking how college was. I tell her I'm in the middle of my maths and ask her if I can get back to this.

I continue and someone else calls on the landline. I have to unplug my internet connection and answer the phone. Wrong number.

I get back, save what I've done in a .txt file and re-start my PC (my net connection crashes when I unplug from the wall without disconnecting through AOL first).

I re-boot, and my PC crashes, I finally get onto VBF and post this.

Arghh, murphys law at its worst, eh?