Intersecting Chords in a Circle

CAT 2019 Slot 1 · QA · Medium · Circles

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In a circle of radius 11 cm11\text{ cm}, CDCD is a diameter and ABAB is a chord of length 20.5 cm20.5\text{ cm}. If ABAB and CDCD intersect at a point EE inside the circle and CECE has length 7 cm7\text{ cm}, then the difference of the lengths of BEBE and AEAE, in cm\text{cm}, is

  1. A.

    1.5

  2. B.

    3.5

  3. C.

    0.5

  4. D.

    2.5

Answer

C

Explanation

Diameter CD=22 cmCD = 22\text{ cm}. CE=7 cm  ⟹  ED=22−7=15 cmCE = 7\text{ cm} \implies ED = 22 - 7 = 15\text{ cm}. By intersecting chords theorem, AE×BE=CE×ED=7×15=105AE \times BE = CE \times ED = 7 \times 15 = 105. Also AE+BE=20.5AE + BE = 20.5. (BE−AE)2=(AE+BE)2−4(AE×BE)=(20.5)2−4(105)=420.25−420=0.25(BE - AE)^2 = (AE + BE)^2 - 4(AE \times BE) = (20.5)^2 - 4(105) = 420.25 - 420 = 0.25. BE−AE=0.25=0.5 cmBE - AE = \sqrt{0.25} = 0.5\text{ cm}.

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