In the figure above, points P and Q lie on the circle with : GMAT Problem Solving (PS) - Page 2
Check GMAT Club App Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 03 Dec 2016, 07:55

### GMAT 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

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# In the figure above, points P and Q lie on the circle with

Author Message
TAGS:

### Hide Tags

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7060
Location: Pune, India
Followers: 2082

Kudos [?]: 13246 [1] , given: 221

### Show Tags

20 Oct 2013, 19:35
1
KUDOS
Expert's post
obs23 wrote:
chetanojha wrote:
Check another approach for this problem.

Can we assume that coordinates of the center O are (0,0) here, can we always assume so?

From the figure it is clear that O lies at (0, 0). O is the point where x axis and y axis intersect in the figure.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 06 May 2016 Posts: 18 WE: Education (Education) Followers: 0 Kudos [?]: 8 [1] , given: 5 Re: In the figure above, points P and Q lie on the circle with [#permalink] ### Show Tags 16 May 2016, 12:37 1 This post received KUDOS Hello all, I just got this question in my GMATprep practice test, and came looking here for an explanation afterwards. I understand the math in the comments above, and why the answer is what it is. What concerns me, is that the image is not drawn to scale. It did not say the image was not drawn to scale, and this is an official question, so why could I not assume the image was drawn to scale? (When one has time to look at the question properly it is obvious that it is not drawn to scale - but I have up to now just been assuming that geometry questions are drawn to scale unless clearly indicated that they are not.) Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7060 Location: Pune, India Followers: 2082 Kudos [?]: 13246 [1] , given: 221 Re: In the figure above, points P and Q lie on the circle with [#permalink] ### Show Tags 17 May 2016, 00:39 1 This post received KUDOS Expert's post etienneg wrote: Hello all, I just got this question in my GMATprep practice test, and came looking here for an explanation afterwards. I understand the math in the comments above, and why the answer is what it is. What concerns me, is that the image is not drawn to scale. It did not say the image was not drawn to scale, and this is an official question, so why could I not assume the image was drawn to scale? (When one has time to look at the question properly it is obvious that it is not drawn to scale - but I have up to now just been assuming that geometry questions are drawn to scale unless clearly indicated that they are not.) You cannot solve the questions by assuming that the diagram is drawn to scale. Two angles which look equal may not actually be equal. You can just assume the basics - straight lines that look straight are straight; points are in the order in which they are drawn. But don't take calls based on relative length of lines, relative measure of angles according to the diagram drawn etc. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Manager
Joined: 12 Oct 2009
Posts: 115
Followers: 2

Kudos [?]: 62 [0], given: 3

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]

### Show Tags

13 Oct 2009, 04:29
DenisSh wrote:
Sorry, didn't get that part:
Again, using the distance formula, (s+\sqrt{3})^2 + (t-1)^2 = PQ^2 = 8. ->(2)

The method used to find the length of OP or OQ from origin i.e. the distance formula has been used here as well to find the length of PQ with P as (-[square_root]3,1) and Q as (s,t) which gives us
{s - (-[square_root]3)}^2 + {t-1}^2 = PQ^2 = 8 ->equation 2

from Right angle Triangle POQ we had got the length of PQ^2 as 8
Senior Manager
Affiliations: PMP
Joined: 13 Oct 2009
Posts: 312
Followers: 4

Kudos [?]: 158 [0], given: 37

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]

### Show Tags

13 Oct 2009, 12:15

S^2 + T^2 = 4 --> equation 1

(t-1)^2 + (s-(-sqrt(3))^2 = 2^2 + 2^2 = 8 --> equation 2

solving for s gives 1
_________________

Thanks, Sri
-------------------------------
keep uppp...ing the tempo...

Press +1 Kudos, if you think my post gave u a tiny tip

Senior Manager
Joined: 30 Aug 2009
Posts: 286
Location: India
Concentration: General Management
Followers: 3

Kudos [?]: 156 [0], given: 5

### Show Tags

13 Nov 2009, 06:32
IndianGuardian wrote:
How would you solve it by the distance formula kp1811?

from the figure we can see that OP=OQ = radius of the semicircle. Also Angle POQ = 90degree

thus PQ^2 = OP^2 + OQ^2-------eqn1

using distance formula (O is origin)
OP = sqrt [(1-0)^2 + (sqrt -3 -0)^2] = sqrt [4] = 2 = OQ

also OQ = sqrt [ (s-0)^2 + (t-0)^2] = sqrt[s^2 + t^2]
OQ^2 = s^2+t^2 =4 ----eqn2

now PQ = 2sqrt 2 when we put values of OP and OQ in eqn 1

using distance formula length of PQ = sqrt [ (s+ sqrt3)^2 + (t-1)^2]
PQ^2 = (s+sqrt3)^2 + (t-1)^2 = 8--------eqn3

soving eqn 2 and 3 we get sqrt 3 s = t.Equating this in eqn 2 we get s=1
Senior Manager
Joined: 05 Oct 2008
Posts: 273
Followers: 3

Kudos [?]: 367 [0], given: 22

### Show Tags

17 Nov 2009, 22:51
Because the explanation by IndianGuardian seems incorrect:

Lets see how:
First draw a perpendicular from the x-axis to the point P. Lets call the point on the x-axis N. Now we have a right angle triangle \triangle NOP. We are given the co-ordinates of P as (-\sqrt{3}, 1). i.e. ON=\sqrt{3} and PN=1. Also we know angle PNO=90\textdegree. If you notice this is a 30\textdegree-60\textdegree-90\textdegreetriangle with sides 1-sqrt3-2 and angle NOP=30\textdegree.

The coordinates of point P is (-\sqrt{3}, 1) ie ON = 1 and PN = root3 - not the other way round as explained in his posting quoted in blue above.

I get PO = QO = 2 (RADIUS)
POQ is a rt angled Triangle.
Hence, isn't PN = 2 sqrt2 ?
So wouldn't s = 2 sqrt2 - sqrt3

Can someone explain how to solve this. Thanks.
Manager
Joined: 23 Apr 2010
Posts: 135
Location: Tx
Schools: NYU,UCLA,BOOTH,STANFORD
Followers: 1

Kudos [?]: 106 [0], given: 36

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]

### Show Tags

18 Jun 2010, 03:35
I understood bunuel's solution ok but i have a problem with the other one.

I got every step of maths but how did you get (t-1) too. However How did you decide that "t" is bigger than 1. Because it can be less then one then it should became (1-t). Can anyone explain it plz just that part??
Intern
Joined: 27 Jul 2010
Posts: 20
Location: Bangalore
Followers: 0

Kudos [?]: 4 [0], given: 0

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]

### Show Tags

30 Sep 2010, 22:53
DenisSh wrote:

In the figure above, points P and Q lie on the circle with center O. What is the value of s?

This is how I solved it.. This will take less than a minute...

Drop perpendiculars from P and Q. Mark the points as X and Y.

Now, PXO and QYO are right angled triangles.

We know, PO =1 and XO = sqrt(3). So PO = radius = 2.

XOP + POQ + QOY =180
so, QOY = 30,
This gives, t= sqrt (3) and s = 1.
_________________

Nothing is free.. You 've to earn it!!!

Manager
Joined: 15 Apr 2010
Posts: 170
Followers: 4

Kudos [?]: 69 [0], given: 3

### Show Tags

05 Oct 2010, 11:58
OP and OR are prependicular to each other. Hence slope of one must be negative inverse of the other.
Slope of OP = -1/sqrt{3}
Slope of OQ = s/r and it should be equal to sqrt{3}.
Hence s = r * sqrt{3} -----------------(1)

Since OP and OQ represent radius of the circle, their lengths must be equal.
Length of OP = sqrt{(sqrt{3})^2 + 1^2}
Length of OQ = sqrt{r^2 + s^2}

OP = OQ, Also substitue the value of s obtained in (1).
Upon solving, we'll get two values of r +1 and -1. Now since r lies in quad I r has to be +ve. Hence +1.
Manager
Status: Current MBA Student
Joined: 19 Nov 2009
Posts: 127
Concentration: Finance, General Management
GMAT 1: 720 Q49 V40
Followers: 13

Kudos [?]: 340 [0], given: 210

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]

### Show Tags

04 Jan 2011, 10:14
That's exactly what I needed. I fully understand the problem now. Thanks Bunuel!
Manager
Joined: 19 Aug 2010
Posts: 76
Followers: 3

Kudos [?]: 24 [0], given: 2

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]

### Show Tags

13 Feb 2011, 08:40
jullysabat wrote:
BalakumaranP wrote:
DenisSh wrote:

In the figure above, points P and Q lie on the circle with center O. What is the value of s?

This is how I solved it.. This will take less than a minute...

Drop perpendiculars from P and Q. Mark the points as X and Y.

Now, PXO and QYO are right angled triangles.

We know, PO =1 and XO = sqrt(3). So PO = radius = 2.

XOP + POQ + QOY =180
so, QOY = 30,
This gives, t= sqrt (3) and s = 1.

Yeah this is the quickest way I think...
But I think the underlined portion is stated wrongly....

The method with the angles is the quickest one. However, I solved it using the slopes. PO and QO have negative reciprocal slopes. The slope of PO=$$\frac{-1}{\sqrt{3}}$$
The slope QO is the native reciprocal to PO:$$\sqrt{3}$$
This means that we have to solve the system:
$$\frac{t}{s}=\sqrt{3}$$
$$r^2+t^2=4$$

Solve for s and we have $$s=1$$
Director
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 920
Followers: 14

Kudos [?]: 331 [0], given: 123

Re: GMAT prep question - Geometry [#permalink]

### Show Tags

15 Apr 2011, 02:55
The equation of the line PO is y = -1/sqrt(3) x

The slope of the perpendicular line QO is sqrt(3). Hence the equation of line QO => y = sqrt(3) x

One way is finding the intersection of the circle and the line QO. We know that radius = QO = 2

The equation of the circle is x^2 + y^2 = 2^2 ----(1)
y = sqrt(3) x ----(2)

Solving 1 and 2 we get x = 1, -1 and y = sqrt(3) and -sqrt(3)

Hence x = 1 since we are looking in the first quadrant. Answer B
SVP
Joined: 16 Nov 2010
Posts: 1672
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 33

Kudos [?]: 504 [0], given: 36

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]

### Show Tags

15 Apr 2011, 04:27
t/s * -1/root(3) = -1

t/s = root(3)

3 + 1 = t^2 + s^2

=> 4 = 4s^2

=> s = 1 (as s is +ve)

_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7060
Location: Pune, India
Followers: 2082

Kudos [?]: 13246 [0], given: 221

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]

### Show Tags

17 Apr 2011, 16:38
ravsg wrote:
had it been a DS question, was it okay to assume O as origin ??

It is mentioned in the question that it is a circle with center O. If it is obviously the center, it will be given. If there are doubts and it is not mentioned, you cannot assume it. You cannot assume that the figure will be to scale.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Senior Manager
Joined: 08 Nov 2010
Posts: 417
Followers: 7

Kudos [?]: 100 [0], given: 161

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]

### Show Tags

19 Apr 2011, 05:04
I solved it with triangles and not lines.
If we can c that the left triangle have two sides (squrt 3 and 1)
we can immediately know that the hypotenuse = 2 (thats the radii as well)
so we also know very easy that the angels are 30,60,90
and if we know the radii and the angels - we can also know all of that very easy on the right triangle as well - just make sure u put the 30 and 60 deg in the right place.
_________________
VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1353
Followers: 17

Kudos [?]: 236 [0], given: 10

Re: Plane Geometry, Semicircle from GMATPrep [#permalink]

### Show Tags

30 Apr 2011, 19:47
for 30 deg triangle sides are in the ratio 3^(1/2) : 1 : 2
for 60 deg triangle sides are in the ratio 1: 3^(1/2) : 2

Hence S = 1.
_________________

Visit -- http://www.sustainable-sphere.com/
Promote Green Business,Sustainable Living and Green Earth !!

Manager
Joined: 25 Sep 2010
Posts: 85
Schools: HBS, LBS, Wharton, Kelloggs, Booth
Followers: 0

Kudos [?]: 6 [0], given: 15

Re: ps - value of s [#permalink]

### Show Tags

19 Jul 2011, 21:56
s = 1.
According to the diagram,
we see that OP and OQ are perpendicular.
Let us not assume anything here. The figure might be misleading to some because some may think the points are symmetrical. They are not.

Since, OP is perp. to OQ,
slope of OP * slope of OQ = -1
=> (-1/sqrt3)*(t/s) = -1
=> t/s = sqrt3
=> t = s*sqrt3
Hence, point Q can be written as Q(s,s*sqrt3)

Now, since both points are on the circumference of the circle,
distance to the centre(origin here) will be the same.

Distance from (x,y) to (0,0) is sqrt(x^2 + y^2)
=>dist. from (-sqrt3,1) to (0,0) = sqrt(3+1) = 2 ----(1)
similarly dist. from (s,s*sqrt3) to (0,0) = 2s -----(2)

equating (1) and (2), we get s = 1.
Hope it helps.
VP
Joined: 24 Jul 2011
Posts: 1122
GMAT 1: 780 Q51 V48
GRE 1: 1540 Q800 V740
Followers: 120

Kudos [?]: 526 [0], given: 19

### Show Tags

05 Jan 2012, 04:47
This is not a question on reflection. This is a question on finding the coordinates of a given point within some stated conditions. It would have been a reflection question if angle POY was the same as angle QOY. It is not, so this is not a reflection question. Solve it according to the rules explained in my last post.

I took the product of the slopes as -1 because if two lines are perpendicular, the product of their slopes must be -1.
OP and OQ are clearly perpendicular as given in the figure, so their slopes must have a product of -1.
_________________

GyanOne | Top MBA Rankings and MBA Admissions Blog

Premium MBA Essay Review|Best MBA Interview Preparation|Exclusive GMAT coaching

Get a FREE Detailed MBA Profile Evaluation | Call us now +91 98998 31738

Manager
Joined: 29 Jul 2011
Posts: 107
Location: United States
Followers: 5

Kudos [?]: 57 [0], given: 6

### Show Tags

05 Jan 2012, 14:39
You can use an alternate approach here.

NOTE: When you have a right angle at origin O and at center of the circle, both intersection edges of the angle (with the circle) will always have the co-ordinates flipped. that is P(x1, y1) = Q(y1, x1), keeping the quadrants into respect.

Imagine a right angle travelling this way inside the circle and how the edges change with the movement.

Therefore, here P(-sqrt(3), 1) for Q will be (1, sqrt(3)). s = 1.
_________________

I am the master of my fate. I am the captain of my soul.
Please consider giving +1 Kudos if deserved!

DS - If negative answer only, still sufficient. No need to find exact solution.
PS - Always look at the answers first
CR - Read the question stem first, hunt for conclusion
SC - Meaning first, Grammar second
RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

Re: Coordinate Geometry and radius - Gmatprep   [#permalink] 05 Jan 2012, 14:39

Go to page   Previous    1   2   3   4    Next  [ 63 posts ]

Similar topics Replies Last post
Similar
Topics:
In the figure above, what is the distance from point P to point Q? 1 12 May 2015, 12:09
23 In the figure above, showing circle with center O and points 8 28 Feb 2013, 04:23
21 In the figure above, showing circle with center O and points 19 25 Feb 2013, 01:21
1 In the figure above, points P and Q lie on the circle with 1 29 Jan 2012, 11:41
146 In the figure above, point O is the center of the circle and 39 16 Jan 2011, 15:56
Display posts from previous: Sort by