It is currently 23 Mar 2018, 15:19

### 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

# Circle passes through points (1, 2) , (2, 5)

Author Message
Manager
Joined: 14 Jun 2008
Posts: 162
Circle passes through points (1, 2) , (2, 5) [#permalink]

### Show Tags

26 Aug 2008, 09:47
Circle passes through points $$(1, 2)$$ , $$(2, 5)$$ , and $$(5, 4)$$ .

What is the diameter of the circle?
* $$\sqrt{18}$$
* $$\sqrt{20}$$
* $$\sqrt{22}$$
* $$\sqrt{26}$$
* $$\sqrt{30}$$

i see no easy way of solving this based on teh OA.
any thoughts?

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Senior Manager
Joined: 06 Apr 2008
Posts: 401
Re: m21 q20 : solve in 2 minutes? [#permalink]

### Show Tags

26 Aug 2008, 10:02
sset009 wrote:
Circle passes through points $$(1, 2)$$ , $$(2, 5)$$ , and $$(5, 4)$$ .

What is the diameter of the circle?
* $$\sqrt{18}$$
* $$\sqrt{20}$$
* $$\sqrt{22}$$
* $$\sqrt{26}$$
* $$\sqrt{30}$$

i see no easy way of solving this based on teh OA.
any thoughts?

The three points form a right angle triangle at point (2,5) .

Distance between (1,2) and (2,5) = $$sqrt (10)$$ (1,2) and (5,4) = $$sqrt (20)$$ and (2,5) and (5,4) = $$sqrt(10)$$

Therefore diameter of circle = hypoteneuse of triangle = $$sqrt(20)$$

Since any right angle triangle formed between three points of circle hypoteneuse is diameter
Manager
Joined: 04 Sep 2007
Posts: 207
Re: m21 q20 : solve in 2 minutes? [#permalink]

### Show Tags

26 Aug 2008, 14:07
nmohindru wrote:
sset009 wrote:
Circle passes through points $$(1, 2)$$ , $$(2, 5)$$ , and $$(5, 4)$$ .

What is the diameter of the circle?
* $$\sqrt{18}$$
* $$\sqrt{20}$$
* $$\sqrt{22}$$
* $$\sqrt{26}$$
* $$\sqrt{30}$$

i see no easy way of solving this based on teh OA.
any thoughts?

The three points form a right angle triangle at point (2,5) .

Distance between (1,2) and (2,5) = $$sqrt (10)$$ (1,2) and (5,4) = $$sqrt (20)$$ and (2,5) and (5,4) = $$sqrt(10)$$

Therefore diameter of circle = hypoteneuse of triangle = $$sqrt(20)$$

Since any right angle triangle formed between three points of circle hypoteneuse is diameter

nmohindru, Could you please describe how you were able to ascertain that it is a right triangle at (2,5)? I like your approach though.
Manager
Joined: 14 Jun 2008
Posts: 162
Re: m21 q20 : solve in 2 minutes? [#permalink]

### Show Tags

26 Aug 2008, 14:28
yeah nice approach, but how did u spot the 90 degrees?
Intern
Joined: 10 Aug 2008
Posts: 10
Location: Research Triangle Park, NC
Re: m21 q20 : solve in 2 minutes? [#permalink]

### Show Tags

26 Aug 2008, 19:26
sset009 wrote:
yeah nice approach, but how did u spot the 90 degrees?

I think he spotted it by drawing a diagram without the circle. Or not.

Let's start with (1,2) and (2,5). Draw a right triangle keeping the X-coordinate constant, and we have a third point at (1,5). This gives us three legs of a right triangle. Side A=(1,2) to (1,5) = 3, side B=(1,5) to (2,5) = 1, side C=(2,5) to 1,2)=$$sqrt(A^2+B^2)=sqrt(1^2+3^2)=sqrt(10)$$. Use the same method for the points (2,5) and (5,4), but holding the Y-coordinate constant, and the hypotenuse is also $$sqrt(10)$$. Applying the same method for points (5,4) and (1,2) you find that the hypotenuse for that triangle is $$sqrt(20)$$. Looking at the triangle with the lengths of the legs and hypotenuse calculated you can see that the triangle is a right triangle.

cP
_________________

Just a few electrons short of a full cloud...

SVP
Joined: 07 Nov 2007
Posts: 1756
Location: New York
Re: m21 q20 : solve in 2 minutes? [#permalink]

### Show Tags

26 Aug 2008, 19:33
sset009 wrote:
yeah nice approach, but how did u spot the 90 degrees?

if you see the two sides are sqrt(10),sqrt(10) and third side is sqrt (20)
they are in the ratio 1:1:sqrt(2) .. this possible only for right angle traingle.

this large side must be diameter.

Great job!! Mohindru.
_________________

Smiling wins more friends than frowning

Intern
Joined: 03 Mar 2008
Posts: 44
Re: m21 q20 : solve in 2 minutes? [#permalink]

### Show Tags

26 Aug 2008, 19:45
nmohindru wrote:
sset009 wrote:
Circle passes through points $$(1, 2)$$ , $$(2, 5)$$ , and $$(5, 4)$$ .

What is the diameter of the circle?
* $$\sqrt{18}$$
* $$\sqrt{20}$$
* $$\sqrt{22}$$
* $$\sqrt{26}$$
* $$\sqrt{30}$$

i see no easy way of solving this based on teh OA.
any thoughts?

The three points form a right angle triangle at point (2,5) .

Distance between (1,2) and (2,5) = $$sqrt (10)$$ (1,2) and (5,4) = $$sqrt (20)$$ and (2,5) and (5,4) = $$sqrt(10)$$

Therefore diameter of circle = hypoteneuse of triangle = $$sqrt(20)$$

Since any right angle triangle formed between three points of circle hypoteneuse is diameter

Its nice that you were able to spot the hypoteneuse and the point of right angle. However with this approach, you would be stuck if this is not the case.
The following is an approach which would give result no matter where the points are on the circle. The drawback is that it would take 3-5 min to solve.

Equation of circle is

(x-h)^2 + (y-k)^2 = r^2
where (h,k) are coordinates of center, r = radius and (x,y) are points on circle

Insert the 3 points to obtain the following equations

h^2 + k^2 - 2h -4k + 5 = r^2
h^2 + k^2 - 4h -10k + 29 = r^2
h^2 + k^2 - 10h -8k + 41 = r^2

Solve them and (h,k) = (3,3)
Hence r = sqrt 5
Diameter = 2 sqrt 5 = sqrt 20
Manager
Joined: 09 Jul 2007
Posts: 240
Re: m21 q20 : solve in 2 minutes? [#permalink]

### Show Tags

26 Aug 2008, 20:05
On the simira lines, but I did it like this :

Let the coordinate of the center be (x,y ) ; now the distance from all the 3 points will be equal ( radius)

so v[ (x-1)^2 + (y-2)^2] = v[(x-2)^2 + (y-5)^2 ] = v[ (x-5)^2 + y-4)^2 ]

that gives us two equations ( after easy cancellation ) x+3y=12 and 2x+ y= 9 ; which provides us (x,y) = (3,3)

so radius = v [ (3-1)^2 + (3-2)^2 ] = v[ 4] ; so diameter= 2v[4] = v[20]

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Re: m21 q20 : solve in 2 minutes?   [#permalink] 26 Aug 2008, 20:05
Display posts from previous: Sort by

# Circle passes through points (1, 2) , (2, 5)

Moderator: chetan2u

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.