Circumcircle of a triangle

Question

The circumcircle of a triangle passes through all three vertices of the triangle. It's center is located at the point of intersection of the perpendicular bisectors of the sides of the triangle. Find a general form equation of the circumcircle of the triangle with vertices (2,1), (3,6), (5,3)

Answer 1

This is an involved problem that requires you to understand perpendicularity, finding slopes and lines, the distance formula, and of course the standard form of a circle.

 

To begin, you'll need to find the equation of the perpendicular bisector to two of the sides of the triangle. For this example, I'll use the side connecting (2,1) and (5,3).

The midpoint of the side is [ (2+5)/2, (1+3)/2 ] = (7/2, 2)

The slope of this side is (3-1)/(5-2) = 2/3

A line that is perpendicular to this side will have a slope that is the negative reciprocal 2/3 -> -3/2

The perpendicular bisector then has slope -3/2 and passes through the point (7/2, 2).

You can use point-slope form [ y-y₁ = m (x - x₁) ] to find the equation of this line.

y - 1 = -3/2 (x-2)

y - 1 = (-3/2) x + 3

y = (-3/2) x + 4

 

Do the same thing for another of the sides of the triangle, and you'll get another equation of a line.

 

Once you have the two equations, set them equal and solve for x. Plug this into the equation of the line to find the point where the lines intersect. This is the center of your circle.

 

Now you need the radius, which is the distance from your center to any of the three vertices given. (It might be worthwhile to calculate at least two of these distances to make sure they're the same, and your center is correct.) Once you have the radius, plug the center and radius into the general formula of a circle:

(x-h)² + (y-k)² = r²

Where (h,k) is your center, and r is the radius.

 

Hope this helps, if you have questions, please comment.