Chords $\overline{AB}$ and $\overline{CD}$ of a circle meet at $Q$ inside the circle. If $CD = 38$, $AQ = 6$, and $BQ = 12$, then what is the minimum length of $CQ$?
+128400 Intersecting chord theorem :
AQ * BQ = CQ * DQ
Let CQ = x
Then since CD = 38, then DQ = (38 - x)
So......
6 * 12 = x (38 - x)
72 = 38x - x^2 rearrange as
x^2 - 38x + 72 = 0 factor
(x - 36) ( x - 2) = 0
Set each factor to 0 and solve for x and we get that
x = 36 or x = 2
Then the minimum length of CQ = 2
