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Let $\overline{AB}$ and $\overline{CD}$ be chords of a circle, that meet at point $Q$ inside the circle. If $AQ = 6,$ $BQ = 12$, and $CD = 38$, then find the minimum length of $CQ$.

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sqdancefan

9514 1404 393

Answer:

  2

Step-by-step explanation:

The products of chord lengths are the same for the intersecting chords:

  AQ×BQ = CQ×DQ

  6×12 = CQ×(38 -CQ)

This gives a quadratic in CQ:

  CQ² -38CQ +72 = 0 . . . . . write in standard form

  (CQ -2)(CQ -36) = 0 . . . . . factor the quadratic

  CQ = 2 or 36 . . . . . . . values of CQ that make the factors zero

The minimum length of CQ is 2 units. (DQ will be 36.)