The intersecting circles x^2+y^2=100 and (x-21)^2+y^2=289 have a common chord. Find its length??

Ok, we've got a circle centered at the origin with radius 10
and a circle centered at (21,0) with radius 17

From this I can reason out the the chord will be a vertical line, a point where the
y values of the two are the same. So that means that

100 - x^2 = 289 - (x - 21)^2

100 - x^2 = 298 - x^2 + 42x - 441

x^2 cancel each other out

100 = 42x - 152
252 = 42x

x = 6

Checking either equation, when x = 6, y = + or - 8

That would make the length of the chord 16.

x² + y² = 100
y² = 100-x²

(x-21)² + y² = 289
(x-21)² + (100-x²) = 289
(x²-42x+441) + (100-x²) = 289
x = 6

y² = 100-x² = 64
y = ±8

The circles intersect at (6,-8) and (6,8).
Distance from (6,-8) to (6,8) = 16 units

A graph of the two circles reveals they intersect at (6, ±8). The length of the chord is 16.