# 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 cmB. 3√5 cmC. 4√10 cmD. 5√10 cm

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 …

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  (OP) = 13 cm

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

(PB)2 = 169 - 9 = 160

PB = 4√10 cm

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