Q4 of 31 Page 229

⨀(O, r1) and ⨀(O, r2) are such that r1> r2. Chord AB of ⨀(O, r1) touches ⨀(O, r2). Find AB in terms of r1 and r2.


Given that circle (O, r1) and circle (O, r2) are such that r1 > r2.


∴ The circles are concentric.


Let chord AB of circle (O, r1) touches circle (O, r2) at P.


Thus, AB is tangent to circle (O, r2).


∴ OP ⊥ AB and P ∈ AB


Here, P is the foot of the perpendicular drawn from centre O on the chord AB of circle (O, r2).


∴ P is the midpoint of AB.


⇒ AB = 2AP … (1)


Consider right angle ΔOPA,


By Pythagoras Theorem,


⇒ OA2 = AP2 + OP2


⇒ r12 = AP2 + r22


⇒ AP2 = r12 – r22


∴ AP =


From (1),


∴ AB = 2


More from this chapter

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2

A, B are the points on ⨀ (O, r) such that tangents at A and B intersect in P. Prove that OP is the bisector of ∠AOB and PO is the bisector of ∠APB.

 3

A, B are the points on ⨀(O, r) such that tangents at A and B to the circle intersect in P. Show that the circle with as a diameter passes through A and B.

 5

In example 4, if r1 = 41 and r2 = 9, find AB.

 1

P is the point in the exterior of ⨀(O, r) and the tangents from P to the circle touch the circle at X and Y.

(1) Find OP, if r = 12, XP = 5


(2) Find m∠XPO, if m∠XOY = 110


(3) Find r, if OP = 25 and PY = 24


(4) Find m∠XOP, if m∠XPO = 80