# finding the radius of one circle given 3 points and circle's center

• Feb 21st 2010, 01:15 PM
alessandromangione
finding the radius of one circle given 3 points and circle's center
show that the points A=(2,5) B=(6,1) C=(-3,5) lie on circle with center ( -1,2)..what is the radius of this circle

sorry to bother u guys so much..but i truly need ur help
• Feb 21st 2010, 01:23 PM
Quote:

Originally Posted by alessandromangione
show that the points A=(2,5) B=(6,1) C=(-3,5) lie on circle with center ( -1,2)..what is the radius of this circle

sorry to bother u guys so much..but i truly need ur help

hi alessandro,

if the distances from those 3 points to (-1,2) are equal,
then, as they are equidistant from that point,
then (-1,2) can be used as the centre of a circle for which the radius
is the distance from (-1,2) to those 3 points.

Hence, to prove, you can use the distance formula $\displaystyle distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

to calculate the distance from (-1,2) to the 3 points.

This is Pythagoras' theorem in co-ordinate form.

If the distances are the same, the 3 points lie on the circle circumference.
• Mar 15th 2010, 08:55 AM
smig1973
Centre of circle and radius given 3 points on the circumference
The co-ordinates you have supplied in your example offer a very simple solution because 2 of your points fall along the same axis, making the centre of the circle=mean value of the 2 points. However should you need to find the centre and radius given 3 points, non of which fall on the same axis the following equations will be all you need. Having spent quite some time trawling through the equations for this problem I have (with some excell assistance from a colleague) managed to put together a straight forward layout suitable for use in excel and most programable calculators.

GIVEN CO-ORDINATES OF 3 POINTS ANYWHERE ON THE CIRCUMFERENCE OF A CIRCLE (X1,Y1), (X2,Y2) AND (X3,Y3)

THE FOLLOWING EQUATIONS WILL DETERMINE THE CO-ORDINATES OF THE CENTRE (X,Y)

X=((((Y2-X1)/(X2-X1))*((Y3-Y2)/(X3-X2))*(Y1-Y3)+((Y3-Y2)/(X3-X2))*(X1+X2)-((Y2-Y1)/(X2-X1))*(X2+X3))*0.5/(((Y3-Y2)/(X3-X2))-((Y2-Y1)/(X2-X1))))

Y=(-1/((Y2-Y1)/(X2-X1))*(X-((X1+X2)*0.5)))+((Y1+Y2)*0.5)

THE RADIUS CAN NOW BE DETERMINED QUITE SIMPLY AS FOLLOWS

SQRT((X-X1)^2+(Y-Y1)^2))

THIS WILL OF COURSE WORK WITH (X2,Y2) AND (X3,Y3)

YOU MAY FIND IT EASIER TO SUBSTITUTE X & Y WITH E & N SO THAT THE VALUES CAN BE INPUT DIRECTLY INTO THE APPROPRIATE CELLS IN EXCEL WITHOUT HAVING TO SCROLL ACROSS THE PAGE.
• Mar 15th 2010, 09:35 AM
tonio
Quote:

Originally Posted by alessandromangione
show that the points A=(2,5) B=(6,1) C=(-3,5) lie on circle with center ( -1,2)..what is the radius of this circle

sorry to bother u guys so much..but i truly need ur help

If the points A,B,C actually lied on a circle with center at (-1,2) then the distance from any of A,B,C to the center (this is, the supposed circle's radius) would be all the same, yet you can easily check that these three distances are all different. Check again the question and the given information.

Tonio
• Mar 15th 2010, 11:12 AM
Soroban
Hello, alessandromangione!

Quote:

Show that the points: A(2,5), B(6,1), C(-3,5) lie on circle with center ( -1,2).
What is the radius of this circle?

This is a case of TMI ... Too Much Information,
. . and it doesn't make sense.

We need only one point, say $\displaystyle A(2,5)$, and the center $\displaystyle (-1,2)$
. . to determine the radius of the circle.

And $\displaystyle (-1,2)$ is not the center of the circle.
. . For points $\displaystyle A,B,C$, the center is: .$\displaystyle \left(\text{-}\tfrac{1}{2},\;\text{-}\tfrac{3}{2}\right)$

So what is the correct statement of the problem?

• Mar 15th 2010, 11:41 AM
Quote:

Originally Posted by alessandromangione
show that the points A=(2,5) B=(6,1) C=(-3,5) lie on circle with center ( -1,2)..what is the radius of this circle

sorry to bother u guys so much..but i truly need ur help

A=(2,5) and B=(6,1) are equidistant from (-1,-2).
However, this requires C to be (-4,5) or another point that same distance from (-1,-2).
• Mar 15th 2010, 04:04 PM
tonio
Quote:

Originally Posted by Archie Meade
A=(2,5) and B=(6,1) are equidistant from (-1,-2).
However, this requires C to be (-4,5) or another point that same distance from (-1,-2).

Yet the OP wrote clearly (-1,2) and not (-1,-2).

I propose we let the OP check and fix his/her own question , and let us not try to mend what we cannot know.

Tonio
• Mar 15th 2010, 04:10 PM