Q. 495.0( 5 Votes )

In the given figure, O is the center of two concentric circles of radii 5 cm and 3 cm. From an external point P tangents PA and PB are drawn to these circles. If PA = 12 cm then PB is equal to


A. 5 cm

B. 3√5 cm

C. 4√10 cm

D. 5√10 cm

Answer :

In given Figure,


OA AP [Tangent at any point on the circle is perpendicular to the radius through point of contact]


In right - angled OAP,


By Pythagoras Theorem


[i.e. (hypotenuse)2 = (perpendicular)2 + (base)2]


(OP)2 = (OA)2 + (PA)2


Given, PA = 12 cm and OA = radius of outer circle = 5 cm


(OP)2 = (5)2 + (12)2


(OP)2 = 25 + 144 = 136


OP = 13 cm …[1]


Also,


OB BP [Tangent at any point on the circle is perpendicular to the radius through point of contact]


In right - angled OBP,


By Pythagoras Theorem


[i.e. (hypotenuse)2 = (perpendicular)2 + (base)2 ]


(OP)2 = (OB)2 + (PB)2


Now, OB = radius of inner circle = 3 cm


And, from [2] (OP) = 13 cm


(13)2 = (3)2 + (PB)2


(PB)2 = 169 - 9 = 160


PB = 4√10 cm

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