Hi,

 

Does anyone know the easiest way of finding the centre of a circle?

 

Cheers

 

Vijay

If you have a straight edge (unmarked ruler) and a compass, it is easy

to locate the center of a circle. Draw any line L through the circle,

so it intersects the circle at A and B. Then with the compass, find

the perpendicular bisector of AB, which will intersect the circle at C

and D. Clearly, CD is a diameter, so if you bisect CD at point O, O is

the center of the circle.

Finding the Center of a Circle from Three Points

 

Date: 05/22/2000 at 10:18:50

From: Christian Furst

Subject: Center point of circle

 

I have the coordinates of three ordered points on a circle. I want to

find a way to define the circle's center. The purpose is to find a way

to draw the part of the circle that is connecting the three points (by

knowing the center and radius).

 

 

--------------------------------------------------------------------------------

 

 

Date: 05/22/2000 at 15:34:50

From: Doctor Rob

Subject: Re: Center point of circle

 

Thanks for writing to Ask Dr. Math, Christian.

 

Let the equation of the circle be

 

(x-h)^2 + (y-k)^2 = r^2,

 

and substitute the three known points, getting 3 equations in 3

unknowns h, k, and r:

 

(x1-h)^2 + (y1-k)^2 = r^2

(x2-h)^2 + (y2-k)^2 = r^2

(x3-h)^2 + (y3-k)^2 = r^2

 

which you can solve simultaneously. First subtract the third equation

from the other two, thus eliminating r^2, h^2, and k^2. That will

leave you with 2 simultaneous linear equations in h and k to solve.

This you can do as long as the 3 points are not collinear. Then those

values of h and k can be used in the first equation to find the

radius:

 

r = sqrt[(x1-h)^2 + (y1-k)^2].

 

Example: Suppose a circle passes through the points (4,1), (-3,7), and

(5,-2). Then we know that:

 

(h-4)^2 + (k-1)^2 = r^2

(h+3)^2 + (k-7)^2 = r^2

(h-5)^2 + (k+2)^2 = r^2

 

Subtracting the first from the other two, you get:

 

(h+3)^2 - (h-4)^2 + (k-7)^2 - (k-1)^2 = 0,

(h-5)^2 - (h-4)^2 + (k+2)^2 - (k-1)^2 = 0,

 

h^2 + 6h + 9 - h^2 + 8h - 16 + k^2 - 14k + 49 - k^2 + 2k - 1 = 0

h^2 - 10h + 25 - h^2 + 8h - 16 + k^2 + 4k + 4 - k^2 + 2k - 1 = 0

 

14h - 12k + 41 = 0

-2h + 6k + 12 = 0

 

10h + 65 = 0

30h + 125 = 0

 

h = -13/2

k = -25/6

 

Then

 

r = sqrt[(4+13/2)^2 + (1+25/6)^2]

= sqrt[4930]/6

 

Thus the equation of the circle is:

 

(x+13/2)^2 + (y+25/6)^2 = 4930/36

 

Understood?

simple innit :)

Hi,

 

Does anyone know the easiest way of finding the centre of a circle?

 

Cheers

 

Vijay

 

 

its usually in the middle :wack: :duffer:

Hi,

 

Does anyone know the easiest way of finding the centre of a circle?

 

Cheers

 

Vijay

look in the middle

look in the middle

 

fook to slow

Hi,

 

Does anyone know the easiest way of finding the centre of a circle?

 

Cheers

 

Vijay

 

fold it in half...... then in half again (so you have a quarter segment)

open it, and the lines cross at the centre......

 

 

not too easy if you cant fold it though!!

There is a tool, cant remember the name sorry, but its a right angle with a straight edge running right through the center of the angle, this finds the middle if you use it twice. Does this make any sense???

cant you do it with a triangle too....

draw a triangle within the circle, then measure the sides of the triangle, and take a right angle line from the halfway point of each side.....

 

do this 3 times and where those lines cross SHOULD be the centre

use a compass, place the point along the edge of the circle, and make a half circle inside the original circle, then do this again on the opposite side of the original circle, where the 2 new inner half circles meet, is the center

Hi,

 

Does anyone know the easiest way of finding the centre of a circle?

 

Cheers

 

Vijay

 

 

GET SOMEONE ELSE TO DO IT!!!!!

:headvswal :headvswal :headvswal :headvswal :headvswal

:duffer: :duffer: :duffer: :duffer: :duffer:

measure the circumference then divide by pie(ask Macca)

that gives you the diameter, then half this and cut a piece of string that long

tie a pen to one end , place tho otherend on the circumference and draw an arch, do this three times and where they all cross is the centre

easy

Finding the Center of a Circle from Three Points

 

Date: 05/22/2000 at 10:18:50

From: Christian Furst

Subject: Center point of circle

 

I have the coordinates of three ordered points on a circle. I want to

find a way to define the circle's center. The purpose is to find a way

to draw the part of the circle that is connecting the three points (by

knowing the center and radius).

 

 

--------------------------------------------------------------------------------

 

 

Date: 05/22/2000 at 15:34:50

From: Doctor Rob

Subject: Re: Center point of circle

 

Thanks for writing to Ask Dr. Math, Christian.

 

Let the equation of the circle be

 

(x-h)^2 + (y-k)^2 = r^2,

 

and substitute the three known points, getting 3 equations in 3

unknowns h, k, and r:

 

(x1-h)^2 + (y1-k)^2 = r^2

(x2-h)^2 + (y2-k)^2 = r^2

(x3-h)^2 + (y3-k)^2 = r^2

 

which you can solve simultaneously. First subtract the third equation

from the other two, thus eliminating r^2, h^2, and k^2. That will

leave you with 2 simultaneous linear equations in h and k to solve.

This you can do as long as the 3 points are not collinear. Then those

values of h and k can be used in the first equation to find the

radius:

 

r = sqrt[(x1-h)^2 + (y1-k)^2].

 

Example: Suppose a circle passes through the points (4,1), (-3,7), and

(5,-2). Then we know that:

 

(h-4)^2 + (k-1)^2 = r^2

(h+3)^2 + (k-7)^2 = r^2

(h-5)^2 + (k+2)^2 = r^2

 

Subtracting the first from the other two, you get:

 

(h+3)^2 - (h-4)^2 + (k-7)^2 - (k-1)^2 = 0,

(h-5)^2 - (h-4)^2 + (k+2)^2 - (k-1)^2 = 0,

 

h^2 + 6h + 9 - h^2 + 8h - 16 + k^2 - 14k + 49 - k^2 + 2k - 1 = 0

h^2 - 10h + 25 - h^2 + 8h - 16 + k^2 + 4k + 4 - k^2 + 2k - 1 = 0

 

14h - 12k + 41 = 0

-2h + 6k + 12 = 0

 

10h + 65 = 0

30h + 125 = 0

 

h = -13/2

k = -25/6

 

Then

 

r = sqrt[(4+13/2)^2 + (1+25/6)^2]

= sqrt[4930]/6

 

Thus the equation of the circle is:

 

(x+13/2)^2 + (y+25/6)^2 = 4930/36

 

Understood?

 

 

kin el mike he only wants to draw a circle not a maths degree lmao :D

If you know the diameter/radius of circle and a pair of dividers/compass, set them to the radius. Now place one leg directly on the circle, and scribe the other one through the circle. Take the compass off and place pivot on the line you have just scribed. scribe again. Repeat another 4 times. Now join lines together to triangulate the centre. If all measuring has been accurate (unlike this picture lmfao ) your 3 lines should meet at the same point, the centre!! :dance:

 

i think T-Pot has the best way

 

ps. are you going to let us know what you're up to Vijay? :D

If you know the diameter/radius of circle and a pair of dividers/compass, set them to the radius. Now place one leg directly on the circle, and scribe the other one through the circle. Take the compass off and place pivot on the line you have just scribed. scribe again. Repeat another 4 times. Now join lines together to triangulate the centre. If all measuring has been accurate (unlike this picture lmfao ) your 3 lines should meet at the same point, the centre!! :dance:

 

 

 

I dont THINK that will work.......

As the circumference of a circle is 3.14 times the radius, the steps you make will all be euall distance away from each other EXCEPT for the first and last, which will be a timy bit apart...

 

this difference will throw out the acuracy....... I THINK.....

 

 

I believe that method will look EXACTLy like your very professional and wonderful diagram! LOL

Cant you just get a ruler and hold it horizontally and move it down the circle until you are measuring the longest distance and draw a line and do the same in the vertical and where the 2 lines meet is the centre. Is this right?

 

Ivan

I reckon he is trying to share his donught !!!!!

no i reckon he's trying to cheat at pin the tail on the donkey!

I dont THINK that will work.......

As the circumference of a circle is 3.14 times the radius, the steps you make will all be euall distance away from each other EXCEPT for the first and last, which will be a timy bit apart...

 

this difference will throw out the acuracy....... I THINK.....

 

 

I believe that method will look EXACTLy like your very professional and wonderful diagram! LOL

 

 

Don't forget, you are striking a chord, not 'following' that segment of arc. the chord is shorter than the arc, hence it fits exactly 6 times.

 

Might do an AutoCAD drawing if i can be bothered to install it on this pc :cool:

hence it fits exactly 6 times.

 

:cool:

 

 

Hmmm... I'm not convinced Dunk.....

 

I hear what you say about the chord being shorter than the arc, this makes sense to me, it being a straght line...

 

however, the chord is equal to the radius, correct??

and the radius is 3.14 times smaller than the circumference.... (C=pi x R)

 

therefore is cant fit exactly.. surely???

 

 

Can you tell I aint had a drink yet??? LOL

OK you made me bust out AutoCAD. ;) :p

 

You will find this drawing is way too overdimensioned, but that is to show the true accuracy of the method.

 

You are basically creating 6 equal triangles.

 

I am a toolmaker by trade and we have used this method for eons now.

 

HTH.

 

Robert.

 

 

Edit... LOL can you tell I haven't had a drink yet??? :D :D

OK you made me bust out AutoCAD. ;) :p

 

 

 

I am a toolmaker by trade and we have used this method for eons now.

 

HTH.

 

Robert.

 

 

Edit... LOL can you tell I haven't had a drink yet??? :D :D

 

OK Robert, that all make sort of sense, and I can see from the diagram... BUT, I'm either an argumentative bugger, or just a dumbass.... but I am sure that when I used the divider method as a schoolboy, the points didnt meet up...

 

Also, I'm sure that I shouldnt be arguing, as I reakon from your weblink in your sig, you probably have or do work at a college or university, and will therefore blow any arguments I make away (as well as being a toolmaker.....) but what the hell.....

 

Give me a while, and I'll try to get my head round your ideas, or I'll come back and ASK quesions LOL....

 

 

might even have a drink to help me "think" LOL

 

Cheers

:D work your way in from the right-angle :D

Ok, another drawing, now I have shown a segment of the 25mm rad.

 

The length of this rad is 26.1799mm. 6 of these add up too 157.0794mm for the circumference.

 

The circumference of the 50mm diameter circle is 3.14159*50=157.0795 so quite close to the AutoCAD maths.

 

LOL, I have never been to college or university of any kind, but just learned exactly what i needed to know for my job.

 

Take care.

 

Rob.