Question

Find the point in the xy plane which is equidistant from (1,1), (7,1), and (1,9).
Please and thanks.

Answer 1

the desired point will be the center of the circle passing through the three points. So, if that is (h,k), we have

(x-1)^2 + (y-1)^2 = r^2
(x-7)^2 + (y-1)^2 = r^2
(x-1)^2 + (y-9)^2 = r^2

Now, you can solve those to find (h,k) and r, or you can note that the three points form a right triangle, whose hypotenuse is the line joining (1,9) and (7,1).

A right triangle's hypotenuse is the diameter of the circumscribed circle. So, that means that the center of the circle is the midpoint of that segment: (4,5).

Now all we have to do is find the distance from (4,5) to any of the points. The distance from (4,5) to (1,1) is 5, so our circle is

(x-4)^2 + (y-5)^2 = 25

Answer 2

thank you Steve