2788.21 – Circle through (1989, y)

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 $(2, 5)$ .

$m = \dfrac{5 - (-3)}{2 - (-2)} = \dfrac{8}{4} = 2$ .

$y = 2x + b$ . $5 = 2 \times 2 + b$ 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 $n$ .