10 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
GRE ® Registrations 2024
Apply
Apply for GRE® Test now & save 10% with ApplyShop Gift Card | World's most used Admission Test for Graduate & Professional Schools
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