Equation of Circle Passing through 3 points
Your instructions are to find the equation of the circle passing through the points (0,0), (2,2), and (6,4).
The hint given in class is to use the fact that the perpendicular bisector of any chord of a circle passes through the center.
![[ (x-h)^2 + (y-k)^2 = r^2 ]](eqncirc.gif)
You will also need the equation of a circle in standard form.
In simpler steps ...
- Find the slope of the line segment connecting the first pair of points (0,0) and (2,2).
- Find the slope of the line perpendicular to that line segment.
- Find the midpoint of the line segment connecting the first pair of points.
- Find the equation of the line perpendicular to and passing through the midpoint of the line segment connecting the first pair of points.
- Repeat the entire process for another pair of points (2,2) and (6,4) or (0,0) and (6,4).
- Find the intersection of the two lines. This is the center.
- Plug in the center for (h,k) and one of the three points [ (0,0) works nicely ] into the circle equation to find the radius.
- Plug the (h,k) and the radius into the circle equation to get the answer [leave the other point as (x,y) ].