Coordinates of Point P

1. Sep 1, 2007

L²Cc

1. The problem statement, all variables and given/known data
In the coordinate plane, the points F(-2, 1), G(1, 4), and H(4, 1) lie on a circle with center P. What are the coordinate of point P?

2. Relevant equations

3. The attempt at a solution

I tried subtracting the x-values, but my answer would be always wrong. I think I'm approaching the question incorrectly.

2. Sep 1, 2007

rock.freak667

General equation of a circle is $$x^2+y^2+2fx+2gy+c=0$$ where $$(-f,-g)$$ is the centre of the circle(i.e. P) and radius,$$r= \sqrt{f^2+g^2-c}$$

sub the points F,G and H into this equation
and you'll get 3 equations with 3 unknowns

more precisely you should get to solve these equations

$$8f+2g+c=-17$$
$$2f+8g+c=-17$$
$$-4f+2g+c=-5$$

Last edited: Sep 1, 2007
3. Sep 1, 2007

symbolipoint

The three points F, G, H, form a triangle; and the segment bisectors of each side will intersect at the center of the circle which contains F, G, and H. Find the lines for the segment bisectors (you only need two of them) and find their point of intersection. That is the center point of the circle.

How do you find each line? It contains the midpoint of a side and has slope which is negative reciprocal of the side; substitute into formula for equation of a line.

4. Sep 4, 2007

chaoseverlasting

Another way is, let the coordinates of point p be (x,y). Find the point of intersection of the lines PF and GH, and let it be M. Then you can use the property PM*FM=GM*HM to solve for x and y.

5. Sep 5, 2007

atavistic

This will be the circumcentre right??? Then we can find the equations of any 2 perpendicular side bisectors and equate them.

Am i right?

6. Sep 5, 2007

symbolipoint

Messages #3 and #5 express the same concept. Message #2 is nice because it immediately puts the information into simultaneous equations which are fairly easy to solve.