56 Views

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


Team Careers360 9th Jan, 2024
Answer (1)
Team Careers360 23rd Jan, 2024

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.

Know More About

Related Questions

TOEFL ® Registrations 2024
Apply
Accepted by more than 11,000 universities in over 150 countries worldwide
Manipal Online M.Com Admissions
Apply
Apply for Online M.Com from Manipal University
View All Application Forms

Download the Careers360 App on your Android phone

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

150M+ Students
30,000+ Colleges
500+ Exams
1500+ E-books