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PA and PB are two tangents drawn from an external point P to a circle with Centre O and radius 4 cm if PA perpendicular PB find th

By Allison

PA and PB are two tangents drawn from an external point P to a circle with Centre O and radius 4 cm if PA perpendicular PB find the length of Each tangent​

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Allison

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Hearing by David Bouchard


As you walk along the ocean

And pee

In which quadsaut or which axsis do the following points like
a.(5,0)
b.(2-3)
c.(3-2)
d.(0-3)
e.(-5-6)​

2 thoughts on “PA and PB are two tangents drawn from an external point P to a circle with Centre O and radius 4 cm if PA perpendicular PB find th”

  1. Diagram:-

    [tex]\setlength{\unitlength}{1.1mm}\begin{picture}(50,55)\thicklines\qbezier(25.000,10.000)(33.284,10.000)(39.142,15.858)\qbezier(39.142,15.858)(45.000,21.716)(45.000,30.000)\qbezier(45.000,30.000)(45.000,38.284)(39.142,44.142)\qbezier(39.142,44.142)(33.284,50.000)(25.000,50.000)\qbezier(25.000,50.000)(16.716,50.000)(10.858,44.142)\qbezier(10.858,44.142)( 5.000,38.284)( 5.000,30.000)\qbezier( 5.000,30.000)( 5.000,21.716)(10.858,15.858)\qbezier(10.858,15.858)(16.716,10.000)(25.000,10.000)\put(25,30){\line(1,1){14}}\put(25,30){\line(1, – 1){14}}\put(39,15.8){\line(3,1){42}}\put(39,44.2){\line(3, – 1){42}}\put(39,16){\line(0,1){28}}\put(25,30){\circle*{1}}\put(81,30){\circle*{1}}\put(22,30){\sf O}\put(40,46){\sf A}\put(40,12){\sf B}\put(83,30){\sf P}\end{picture}[/tex]

    Given:

    • OA=OB=4cm (Radius)
    • ∠APB=90°
    • AP & PB=Tangents

    To find:

    • Length of Tangents

    Solution:

    ⇒ ∠PAO=∠PBO=90°(angle made by Tangent with radius)

    Now,

    ⇒ ∠APB+∠PBO+∠PAO+∠AOB=360°(Sum of all ∠ of quadrilateral)

    ⇒ 90°+90°+90°+∠AOB=360°

    ⇒ 270°+∠AOB=360°

    ⇒ ∠AOB=360°-270°

    ⇒ ∠AOB=90°

    Now join AB in diagram

    ~Applying Pythagoras theorem in ∆AOB

    ⇒ AO²+OB²=AB²

    ⇒ (4cm)²+(4cm)²=AB²

    ⇒ 16cm²+16cm²=AB²

    ⇒ 32cm²=AB

    ⇒ √32cm²=AB

    ⇒ 4√2cm=AB

    Now assume length of each tangent be x

    Since, PAB is a right angled ∆, we can apply Pythagoras theorem in it.

    ~Applying Pythagoras theorem in ∆PAB

    ⇒ PA²+PB²=(4√2cm)²

    ⇒ x²+x²=32cm²

    ⇒ 2x²=32cm²

    ⇒ x²=32cm²/2

    ⇒ x²=16cm²

    ⇒ x=√16cm²

    ⇒ x=4cm

    So the required length of tangent is 4cm.

    Reply

  2. Answer:

    CA is perpendicular to AP and CB is perpendicular to BP

    Again AC = BC = 4 (radius of the circle)

    Also AP = PB = (Tangents from point P)

    So, BPAC is a square.

    => AP = PB = BC = CA = 4 cm

    So, length of tangents are 4 cm each

    I hope this helps!

    Reply

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