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

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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