BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Geometry

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 304
Joined: Wed Jan 27, 2010 8:35 am
Location: International Space Station
Thanked: 11 times
Followed by:3 members

Geometry

by Aman verma » Tue Jun 08, 2010 9:05 am
Q: Two Circles of radius 1 cm touch at point P . A third circle is drawn through the points A , B and C such that PA is the diameter of the first circle and BC perpendicular to AP is the diameter of the second circle . The radius of third circle is

a) 9/5 cm

b) 7/4 cm

c) 5/3 cm

d) ( 10^1/2)/2 cm

e)2 cm
800. Arjun's-Bird-Eye
Top
Source: — Problem Solving |
Legendary Member
Posts: 809
Joined: Wed Mar 24, 2010 10:10 pm
Thanked: 50 times
Followed by:4 members

by akhpad » Tue Jun 08, 2010 10:24 am
Let the center of third circle be O and radius be r.

OA = OB = OC = r
Mid point of BC is D

DC = 1 cm, OD = 3-r

OC^2 = OD^2 + DC^2

r*r = 1 + (3-r)^2

r = 5/3 cm

Answer: C
Top
User avatar
Newbie | Next Rank: 10 Posts
Posts: 5
Joined: Mon Jun 07, 2010 10:15 am
GMAT Score:710

by nmerchant » Tue Jun 08, 2010 10:46 am
Okay it took me quite some time to get this one. But I did.
The radius of the third circle is the radius of the circumcenter of triangle ABC. Circumcenter is the point where the perpendicular bisectors of all three sides meet.

Taking A as the origin, I got B as (3,1) and C as (3, -1). The perpendicular bisector of BC is the x axis (y = 0).
So now I just need to find the intersection of y= 0 and the perpendicular bisector of AB or AC.
Equation of AB is y = mx + c
Since A is the origin, the y intercept c = 0. using B(3.1) I get the slope as 1/3.
Slope of perpendicular bisector of AB is -3 and midpoint of AB (1.5, 0.5) is a point on this line.
Equation of perpendicular bisector of AB is y - 0.5 = -3 (x - 1.5)
y = -3x + 5
Now solving y= 0 with this equation you get x = 5/3 which gives us the circumcenter (5/3, 0)
And radius of the third circle is distance from A (0,0) to the circumcenter (5/3,0) which is 5/3

I wouldn't attempt this in the exam. :) Takes too much time.
Top
Legendary Member
Posts: 809
Joined: Wed Mar 24, 2010 10:10 pm
Thanked: 50 times
Followed by:4 members

by akhpad » Tue Jun 08, 2010 10:51 am
@nmerchant - your approach is not advisable on real GMAT. Please look my post, above.
Top
User avatar
Newbie | Next Rank: 10 Posts
Posts: 5
Joined: Mon Jun 07, 2010 10:15 am
GMAT Score:710

by nmerchant » Tue Jun 08, 2010 11:03 am
Genius! Very efficient approach for the GMAT thanks a ton.
Top
Master | Next Rank: 500 Posts
Posts: 304
Joined: Wed Jan 27, 2010 8:35 am
Location: International Space Station
Thanked: 11 times
Followed by:3 members

by Aman verma » Thu Jun 10, 2010 8:47 am
OAC
800. Arjun's-Bird-Eye
Top
Senior | Next Rank: 100 Posts
Posts: 56
Joined: Tue Jan 13, 2009 12:59 pm
Location: Toronto
Thanked: 1 times

by odod » Thu Jun 10, 2010 11:25 am
Hello - can someone draw a diagram for me? I'm having diffuculty conceptualizing the 3 circles and a diagram would help me figure out where I'm going wrong.

Thanks!
ODOD
Top
Post new topic Post Reply
• Page 1 of 1