Finding the Center of a Circle Using Perpendicular Bisectors

[L²Cc]

Homework Statement

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?

Homework Equations

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.

 

[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

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

 

[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.

 

[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.

 

[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?

 

[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.

About #3 and #5:
The perpendicular bisectors of the sides of a triangle are concurrent (meaning they intersect) at a point equidistant from the vertices. This also means that the vertices lie on a circle whose center is that point of concurrency.