GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
In the above figure, the radii of the concentric circles are 2 and 5 r
[#permalink]
21 Jan 2025, 14:47
21 Jan 2025, 14:47
Expert Reply
00:00
Question Stats:
99% (03:08) correct
0% (00:00) wrong based on 7 sessions
Hide Show timer Statistics
Attachment:
GRE circle.png [ 30 KiB | Viewed 647 times ]
In the above figure, the radii of the concentric circles are 2 and 5 respectively. If the tangent to the smaller circle intersects the bigger circle at the points ' $P$ ' and ' $Q$ ', then what is the length of $P Q$ ?
(A) $\(\sqrt{21}\)$
(B) $\(2 \sqrt{21}\)$
(C) $\(3 \sqrt{21}\)$
(D) $\(4 \sqrt{21}\)$
(E) $\(5 \sqrt{21}\)$
- Part of the project: GRE Quant & Verbal ADVANCED Daily Challenge 2024 Edition - Gain 20 Kudos & Get FREE Access to GRE Prep Club TESTS
- Also replying to the unanswered questions
Re: In the above figure, the radii of the concentric circles are 2 and 5 r
[#permalink]
02 Mar 2025, 13:41
02 Mar 2025, 13:41
Expert Reply
We have assumed the centre of the two concentric circles as O and have joined O with P and T , where OT is perpendicular on PQ (Radius is always perpendicular on tangent).
We know $\(\mathrm{OT}=2\)$ and $\(\mathrm{OP}=5\)$ are the radii of the two circles.
Now, applying Pythagoras theorem i.e. Hypotenuse $\({ }^2=\)$ Perpendicular $\(^2+\)$ Base $\(^2\)$ in the triangle POT, we get $\(\mathrm{PT}=\sqrt{\mathrm{PO}^2-\mathrm{OT}^2}=\sqrt{5^2-2^2}=\sqrt{2 \mathrm{I\)$
Note: - Perpendicular drawn from the centre of the circle to the chord always bisects the chord and as PQ is a chord to the bigger circle and OT is perpendicular drawn from the centre of the circle, we get $\(\mathrm{PT}=\mathrm{TQ}\)$
Thus we get $\(P Q=2(P T)=2 \sqrt{21}\)$, so the answer is $\((B)\)$.
We know $\(\mathrm{OT}=2\)$ and $\(\mathrm{OP}=5\)$ are the radii of the two circles.
Now, applying Pythagoras theorem i.e. Hypotenuse $\({ }^2=\)$ Perpendicular $\(^2+\)$ Base $\(^2\)$ in the triangle POT, we get $\(\mathrm{PT}=\sqrt{\mathrm{PO}^2-\mathrm{OT}^2}=\sqrt{5^2-2^2}=\sqrt{2 \mathrm{I\)$
Note: - Perpendicular drawn from the centre of the circle to the chord always bisects the chord and as PQ is a chord to the bigger circle and OT is perpendicular drawn from the centre of the circle, we get $\(\mathrm{PT}=\mathrm{TQ}\)$
Thus we get $\(P Q=2(P T)=2 \sqrt{21}\)$, so the answer is $\((B)\)$.
gmatclubot
Re: In the above figure, the radii of the concentric circles are 2 and 5 r [#permalink]
02 Mar 2025, 13:41
Moderators:
