The end points of a diameter of a circle are (3, 9) and (11, 3). A triangle inscribed in this circle has two of its vertices at the given points. Find the coordinates of all points at which a third vertex of the triangle can be located for this triangle to have its maximum possible area.

Since all such triangles have the same base, the way to maximize the area is to maximize the altitude, namely by using the radius (OP).

We find the radius OA to be 5, using the distance formula. Thus OP = 5.

The slope of AB is (9 – 3) / (3 – 11) = -6/8 =

-3/4. Hence, the slope of OP is 4/3.

We could find the equation of OP and use the distance formula again to find P’s coordinates; but we can also consider the right trangle OCP whose legs (due to the slope of OP) are 3 and 4: OC = 3, CP = 4. Using the midpoint formula, the coordinates of O are (7, 6), so C is (10, 6) and P is (10, 10).

Using slope again, we find Q is (4, 2).

So the two points are P(10, 10) and Q(4, 2).