In a rectangular coordinate plane, any circle that passes through (-2, -3) and
(2, 5) cannot also pass through (1989, y). Find the value of y.
This is about the fact that no three points on a circle are collinear.
We find the equation, y = mx + b, of the line containing (-2, -3) and (2, 5).
| m = | 5 – (-3) | = | 8 | = 2 |
| 2 – (-2) | 4 |
y = 2x + b. Then 5 = 2 · 2 + b → b = 1 → y = 2x + 1.
y = 2(1989) + 1 = 3979.
So a circle that passes through (-2, -3) and (2, 5) cannot also pass through (1989, 3979).
This problem was probably used in the year 1989. It could be adapted for the current year, of course, or generalized for the year n.