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Circle passes through points (1, 2) , (2, 5) [#permalink]
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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 repost 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.



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Re: m21 q20 : solve in 2 minutes? [#permalink]
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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



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Re: m21 q20 : solve in 2 minutes? [#permalink]
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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.



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Re: m21 q20 : solve in 2 minutes? [#permalink]
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26 Aug 2008, 14:28
yeah nice approach, but how did u spot the 90 degrees?



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Re: m21 q20 : solve in 2 minutes? [#permalink]
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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 Xcoordinate 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 Ycoordinate 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
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Re: m21 q20 : solve in 2 minutes? [#permalink]
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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.
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Re: m21 q20 : solve in 2 minutes? [#permalink]
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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 35 min to solve. Equation of circle is (xh)^2 + (yk)^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



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Re: m21 q20 : solve in 2 minutes? [#permalink]
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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[ (x1)^2 + (y2)^2] = v[(x2)^2 + (y5)^2 ] = v[ (x5)^2 + y4)^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 [ (31)^2 + (32)^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 repost 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?
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