In the figure,two circles with centres A and B of 5 cm and 3 cm touch other internally.If the perpendicular bisector of segment AB meets the bigger circle in P and Q.Find the length of PQ.


As the 2 circles touch internally, 

∴AB = r1 - r2 = 2 cm

Also, PQ is the perpendicular bisector of AB.

∴AC = BC = 1 cm

In right ΔACP,

AP2 = AC2 + CP2

⇒52 = 12 + CP2

⇒CP2 = 24

⇒CP = 2√6 cm

Hence, PQ = 2CP

 = 4√6 cm

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If Two Circles Touch Each Other Internally Then Distance Between There Centres Is The Difference Of Their Radii

AB = 5-3 = 2cm

Also The common chord PQ Is The Perpendicular Bisector Of AB

So The Centre Of AB Is C.

So AC = CB = 1 cm

iN right triangle acp we have

AP2 = AC2+CP2

CP  = Root 24

PQ  = 2CP = 2*Root 24 = 4*Root 6

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