Question : Two circles with centres A and B of radii 6 cm and 4 cm, respectively, touch each other internally. If the perpendicular bisector of AB meets the bigger circle in P and Q, then the value of PQ is:
Option 1: $\sqrt{5}$ cm
Option 2: $2 \sqrt{35}$ cm
Option 3: $\sqrt{35} $ cm
Option 4: $2 \sqrt{5}$ cm
Correct Answer: $2 \sqrt{35}$ cm
Solution :
AB = 6 – 4 = 2 cm
The perpendicular bisector of AB meets the bigger circle in P and Q.
The perpendicular chord bisector equals the radius of a circle.
The perpendicular line bisects the chord.
The perpendicular bisector bisects chord AB at C.
AC = $\frac{AB}{2}$ = $\frac{2}{2}$ = 1 cm
In a $\triangle ACB$, using Pythagoras's theorem,
$(AP)^2=(AC)^2+(PC)^2$
⇒ $(PC)^2=6^2–1^2$
⇒ $(PC)^2=36–1$
⇒ $(PC)^2=35$
⇒ $PC=\sqrt{35}$ cm
The length of PQ = 2 PC = $2\sqrt{35}$ cm.
Hence, the correct answer is $2\sqrt{35}$ cm.
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