In a rectangular coordinate plane, any circle which passes through (-2,-3) and (2,5) cannot also pass through (1989,y). What's the value of y?

## Solution

This is about the fact that no three points on a circle can be collinear.

So we find the equation, $y = mx + b$, of the line containing (-2,-3) and (2,5).

$m = (5 - (-3))/(2 - (-2) = 8/4 = 2$.

$y = 2x + b$. $5 = 2x2 + b$ so $b = 1$ and $y = 2x + 1$.

$y = 2(1989) + 1 = 3979$.

Note: this problem was probably first used in the year 1989. It could be adapted for the current year, of course, or generalized for the year n.