Q29 of 34 Page 1

The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle and BD is tangent to the smaller circle touching it at D and intersecting the larger circle at P, on producing. Find the length of AP.

Given: The radii of two concentric circles with same center as O are 13 cm and 8 cm. AB is a diameter of the bigger circle and BD is tangent to the smaller circle touching it at D and intersecting the larger circle at P, on producing


To Find: length of AP



Construction : Join OD


Clearly OB ⏊ AC [ As tangent to at any point on the circle is perpendicular to the radius through point of contact]


So OBC is a right-angled triangle


So, it will satisfy Pythagoras theorem [ i.e. (base)2 + (perpendicular)2 = (hypotenuse)2 ]


i.e.


(OD)2 + (BD)2 = (OB)2


As OD and OB are the radii of circles, Therefore, OD = 8 cm, OB = 13 cm


⇒ (8)2 + (BD)2 = (13)2


⇒ 64 + (BD)2 = 169


⇒ (BD)2 = 169 - 64 = 105


⇒ BC = √105 cm


Also


PD = BD [ Perpendicular through the center to a chord in a circle bisects the chord]


And


BP = PD + BD


= BD + BD = 2BD = 2√105 cm


Also,


∠APB = 90° [Angle in a semicircle is a right-angle]


So, By Pythagoras theorem [ i.e. (base)2 + (perpendicular)2 = (hypotenuse)2 ]


i.e.


(AP)2 + (BP)2 = (AB)2


Also, AB = 2 × radius of bigger circle = 2(13) = 26 cm


Putting values, we get


(AP)2 + (2√105)2 = (26)2


⇒ (AP)2 + 420 = 676


⇒ (AP)2 = 256


⇒ AP = 16 cm


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