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    CAT Some Results on Chord and Circle - Practice Questions & MCQ

    Edited By admin | Updated on Oct 04, 2023 04:20 PM | #CAT

    Quick Facts

    • 4 Questions around this concept.

    Solve by difficulty

    Two chords AB and CD of a circle intersect each other at P outside the circle. If AB = 2 cm, BP = 6 cm, and CP = 2.5 cm. Then PD is:

    Concepts Covered - 1

    Some Results on Chord and Circle

    Result 1: If two chords of a circle intersect each other, then the products of the lengths of their segments are equal.

    Let us prove this

    Case 1: Let the two chords intersect internally (inside the circle).

    Let AB and CD are two chords intersecting at point P in the circle, and we have to prove (PA) (PB) = (PC) (PD).

    Join AC and BD.

    In triangles APC and PDB,

    ∠APC = ∠DPB                   (Vertically opposite angels)

    ∠CAP = ∠CDB                   (Angles made by arc BC in the same segment)

    ΔAPC is similar to ΔDPB.    (by the AA Similarity Property)

    \mathrm{\therefore \frac{P A}{P D}=\frac{P C}{P B} \Rightarrow(P A)(P B)=(P C)(P D)}

    Case 2: Let the two chords intersect externally (outside the circle).

    Let two chords BA and CD intersect at point P which lies outside the circle, and we have to prove (PA) (PB) = (PC) (PD).

    Join AC and BD.  

    In triangles PAC and PDB,

           ∠PAC = ∠PDB

    and ∠PCA = ∠PBD

    (An external angle of a cyclic quadrilateral is equal to the interior angle at the opposite vertex)

    ΔPAC is similar to ΔPDB.    (by the AA Similarity Property)

    \mathrm{\therefore \frac{P A}{P D}=\frac{P C}{P B} \Rightarrow(P A)(P B)=(P C)(P D)}

    Result 2: The converse of the above Result is also true, i.e., if two line segments AB and CD intersect at P and (PA)(PB) = (PC)(PD), then the four points are concyclic.

    Result 3: If one of the secants (say PCD) is rotated around P so that it becomes a tangent, i.e., points C and D say at T, coincide. We get the following result:

    If PAB is a secant to a circle intersecting the circle at A and B and PT is the tangent drawn from P to the circle, then PA · PB = PT2

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