Circles point pass through

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November 7th 2005, 10:53 AM
stormy_girl

Circles ( Plz help )

A cirlcle passes through A (3;1) B(8;2) C (2.6)
a) Find the point of intersection of the perpendicular bisectors of AB and BC.
b) Find the equation of the circle
January 4th 2006, 12:45 AM
earboth

Hi, it's a little late, but I hope it'll help even now:

To a):
All points P(x,y) on a circle with the center $M(x_M,y_M)$ will be described by the equation
$ (x-x_{M})^2+(y-y_{M})^2=r^2 $

With 3 points you generate 3 equations by insertig the coordinates of the points:
$ A: (3-x_{M})^2+(1-y_{M})^2=r^2 \\ $
$ B: (8-x_{M})^2+(2-y_{M})^2=r^2 \\ $
$ C: (2-x_{M})^2+(6-y_{M})^2=r^2 $
Expand these equations and you get:
$ A: 9-6 x_{M}+(x_{M})^2+1-2y_{M}+(y-{m})^2=r^2 \\ $
$ B: 64-16 x_{M}+(x_{M})^2+4-4y_{M}+(y_{M})^2=r^2 \\ $
$ C: 4-4 x_{M}+(x_{M})^2+36-12y_{M}+(y_{M})^2=r^2 $
Substract (vertically!):
$ A-B: -55+10 x_{M}-3+2y_{M}=0 \\ $
$ A-C: 5-2 x_{M}-35+10y_{M}=0 $
So you've got a system of linear equations:
$ A-B: 10 x_{M}+2y_{M}=58 \\ $
$ A-C: -2 x_{M}+10y_{M}=30 $
which delivers the solution
$ x_{M}=5 &\ y_{M}=4 $
Insert these solutions into equation A:
$ A: (3-5)^2+(1-4)^2=r^2 \\ $
$ A: 4+9=r^2\\ $
$ r=\sqrt{13} $

To b)
Insert all your result into the equation mentioned above:
$ (x-5)^2+(y-4)^2=13 $
That's the equation of the circle.
January 4th 2006, 04:05 AM
ticbol

Quote:

Originally Posted by stormy_girl

A cirlcle passes through A (3;1) B(8;2) C (2.6)
a) Find the point of intersection of the perpendicular bisectors of AB and BC.
b) Find the equation of the circle

Here is one way.

a) The point of intersection of the perpendicular bisectors of AB and BC.

Let us call
line U = perpendicular bisector of AB
line V = perpendicular bisector of BC

Midpoint of AB = ((3+8)/2,(1+2)/2) = (5.5,1.5)

Slope of AB = (2-1)/(8-3) = 1/5
So, slope of line U = -5/1 = -5

Point-slope form of the equation of line U is
(y -1.5) = -5(x -5.5)
Isolating the y,
y = -5(x -5.5) +1.5 ---------(i)

Midpoint of BC = ((8+2)/2,(2+6)/2) = (5,4)

Slope of BC = (6-2)/(2-8) = 4/(-6) = -2/3
So, slope of line V = -(-3/2) = 3/2

Point-slope form of the equation of line V is
(y -4) = (3/2)(x -5)
Isolating the y,
y = (3/2)(x -5) +4 ---------(ii)

At the intersection of lines U and V, their coordinates are the same, so,
y from (i) = y from (ii)
-5(x -5.5) +1.5 = (3/2)(x -5) +4
Clear the fraction, multiply both sides by 2,
-10(x -5.5) +3 = 3(x -5) +8
-10x +55 +3 = 3x -15 +8
55 +3 +15 -8 = 3x +10x
65 = 13x
x = 65/13 = 5 ----------x-coordinate of the intersection of U and V.
Substitute that into, say, (ii),
y = (3/2)(5 -5) +4 = 0+4 = 4 ------y-coordinate of the intersection.

Therefore, the intersection point is (5,4). ---------answer.

-------------------------------------
b) The equation of the circle.

The standard (or is it "general"?) form of the equation of the circle whose centerpoint is at (h,k), and whose radius is r, is:
(x-h)^2 +(y-k)^2 = r^2 ----------***

The radius of this circle is the distance from (5,4) to any of A, B, or C.
Say, we use radius to point A.
By Pythagorean theorem,
r^2 = (5-3)^2 +(4-1)^2
r^2 = 4 +9 = 13

Therefore, the equation of the circle here is
(x-5)^2 +(y-4)^2 = 13 -----------answer.

----------
Ooopps, sorry, I have not established yet that (5,4) is the center of the circle.

Any 3 non-collinear points, at least one point is not collinear, determines a circle.
The intersection point of the perpendicular bisectors of the 3 sides of the triangle determines the center of the circle.

There. So, (5,4) is the center of the circle here---even if this point is shown as the intersection of the perpendicular bisectors of only two sides of triangle ABC.

For exercise, try to solve for the distances from (5,4) to A, B, and C. All three should be equal, all are radii of the circle.