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# m21 #20

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m21 #20 [#permalink]  27 Feb 2009, 10:40
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Question Stats:

79% (01:53) correct 20% (02:12) wrong based on 91 sessions
If a circle passes through points (1, 2) , (2, 5) , and (5, 4) , what is the diameter of the circle?

(A) \sqrt{18}
(B) \sqrt{20}
(C) \sqrt{22}
(D) \sqrt{26}
(E) \sqrt{30}

Source: GMAT Club Tests - hardest GMAT questions

SOLUTION: m21-76117.html#p1224302

[Reveal] Spoiler:
Would someone please explain the answer to this question to me? I know you need to create a square and find point 4,1 to finish the square-got that. The lengths of the sides are therefore 3, no? Therefore, why is the answer the square root of 20 opposed to the square root of 18? How to you arrive at sides 4 and 2 for a square? I'm obviously missing something here. Thank you.
[Reveal] Spoiler: OA
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Re: m21 #20 [#permalink]  28 Feb 2009, 22:23
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Good Q again.Pls check. are the co-ordinates allright?
I Left my fav subject maths almost 12 yrs back
anyway,
using basic equations and solving for (h,k)

(x-h)^2+(y-k)^2=r^2
getting h+3k=12------(1);3h-k=6--------(2)
hence h=3;k=3

d=2r=2(5)^2

dczuchta wrote:
Circle passes through points 1,2 , 2,5 and 5,4 . What is the diameter of the circle?

_____________________________

Would someone please explain the answer to this question to me? I know you need to create a square and find point 4,1 to finish the square-got that. The lengths of the sides are therefore 3, no? Therefore, why is the answer the square root of 20 opposed to the square root of 18? How to you arrive at sides 4 and 2 for a square? I'm obviously missing something here. Thank you.

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Last edited by nitya34 on 28 Feb 2009, 22:30, edited 1 time in total.
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Re: m21 #20 [#permalink]  28 Feb 2009, 22:28
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Expert's post
Kindly note that you can write math expressions in a full-blown way if you enclose your math expression in [m]...[/m] tags

Example:

[m]MATH EXPRESSION[/m] Tag

P.S. Do not Disable BBCode
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Re: m21 #20 [#permalink]  28 Feb 2009, 23:24
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There are a few "rules" or "shortcuts" that can really help you out with this type of a question.

First rule to note, when you have a triangle inscribed inside a circle, if one of the 3 angles is a right angle, the line opposite that right angle will be the diameter of the circle.

How do we know here that we have a triangle with a 90 degree corner? If we have the equation of a line, such as y = 2x + 4, how do we represent the perpendicular to that line? we flip and negate the slope. So a slope of 2x would become -1/2x.

Look at the lines we have here. I will label the points with A, B, and C. A = {1,2} B = {2,5} C = {5,4}

Since we know that perpendicular lines will form a 90 degree angle and we only need to look at the slope to find perpendicular lines, we'll only find the slopes of each of these lines.

Line AB slope = \frac{(5-2)}{(2-1)} = 3.

Line BC Slope = \frac{(4-5)}{(5-2)} = -1/3.

Since we immediately see an inverse, negative slope value, we know that these two lines form a perpendicular and the angle created will be 90 degrees. This also means that since the question tells us it is inscribed in a circle, the line directly opposite the 90 degree angle will be the diameter of the circle. So, Angle ABC = 90 degrees, and Line AC = diameter.

Now to find the diameter, we must treat line AC as the hypotnuse and use the pythagorean theorum to find it's length, which will also be the diameter.

Take (C_x - A_x)^2 + (C_y - A_y)^2 = diameter of the circle. C_x is the X value of point C, or 5.

This is (5 - 1)^2 + (4 - 2)^2 = 20 so the answer = sqrt{20}

***I just realized that the triangle I drew is correct but the labels along the X-axis are incorrect. The red lines are the y-axis and x-axis, but along the X axis, I actually labelled the X at 1. The points are correct, the label is not.***

Attachment:

DiameterOfCircle.jpg [ 77.58 KiB | Viewed 6642 times ]

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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. Director Joined: 04 Jan 2008 Posts: 919 Followers: 36 Kudos [?]: 129 [0], given: 17 Re: m21 #20 [#permalink] 28 Feb 2009, 23:47 Thanks.Got it.Thats the best approach +1 to you. Btw,Whats wrong in the equation approach I applied? Am i doing some silly mistake?becoz i am not getting the result _________________ SVP Joined: 30 Apr 2008 Posts: 1893 Location: Oklahoma City Schools: Hard Knocks Followers: 27 Kudos [?]: 398 [1] , given: 32 Re: m21 #20 [#permalink] 01 Mar 2009, 06:31 1 This post received KUDOS There is nothing wrong with your equation, you are using it correctly, however, the equation we choose to use must be the right equation to find the information we are missing and need. We might be able to use this equation and find some of the missing information {h,k}, but here we don't necessarily need that information to answer the question. Here, you used the equation perfectly. The equation, when used to find the value of {h,k} will tell you the coordinates of the radius. You found that the radius is located at {3,3}, and if you look at the picture I drew (ignoring the x-axis labeling) you'll see that {3,3} is indeed the midpoint between the {1,2} point and {5,4} point (I labeled A and C}. So, while you found the radius, this was just one of the steps in your process to find the diameter. In your equation of d=2r, I'm not sure how the 2(5)^2 works. Also, I find it interesting that you found the center of the circle at {3,3}. In order to do this, you would have to know the value of r^2 in order to leave h and k as the remaining variables. If you already knew the value of r^2, then 2 times the square root of r^2 would be the diameter. The equation of a circle is like reverse engineering the pythagorean theorum. You know the radius (hypotnuse) and you know a point on the circle, so you're moving backwards so-to-speak to find what other point would make the forumula complete. The only way this works is if you know the length of the radius. if you know the length of the radius, then you can just double it to find the diameter which is the point of the question. nitya34 wrote: Good Q again.Pls check. are the co-ordinates allright? I Left my fav subject maths almost 12 yrs back anyway, using basic equations and solving for (h,k) (x-h)^2+(y-k)^2=r^2 getting h+3k=12------(1);3h-k=6--------(2) hence h=3;k=3 d=2r=2(5)^2 dczuchta wrote: Circle passes through points 1,2 , 2,5 and 5,4 . What is the diameter of the circle? Answer: Square root of 20. _____________________________ Would someone please explain the answer to this question to me? I know you need to create a square and find point 4,1 to finish the square-got that. The lengths of the sides are therefore 3, no? Therefore, why is the answer the square root of 20 opposed to the square root of 18? How to you arrive at sides 4 and 2 for a square? I'm obviously missing something here. Thank you. _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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Re: m21 #20 [#permalink]  04 May 2010, 07:05
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distance between two points
____________________
√ (x2 - x1)² + (y2 - y1)²
=>
A→B d =√10, B→C d=√10, C→A d = √20 ( from formula above)
we see
AB² + BC² = AC², (√10)² + (√10)² = (√20)² => 10 + 10 = 20 => 20 = 20
so diameter is
AC = √20
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Re: m21 #20 [#permalink]  04 May 2010, 08:15
Guys, correct me if I am wrong in my logic.
Ignore the point (2,5). Considering the fact that the circle touches all the points, the line joining the (1,2)--say point A and (5,4)---say point B is the diameter(AB).
Now, draw perpendiculars from A and B across the coordinates which will join at Point C---(5,2).
Triangle, thus, formed is right angled triangle- triangle ABC.
Using simple coordinate numbers, the length of AC = 4 and length of BC = 2.
Applying Pythogoras Theorem, the diameter i.e. the hypotenuse of the triangle ABC = sqrt (20).

draw the above figure with simple coordinates to follow the above logic.
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Re: m21 #20 [#permalink]  04 May 2010, 13:28
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I did it the same way as deepak4mba, which has got to be the fastest and most intuitive.

Just sketch it out and you'll see that the line from (1,2) to (5,4) is obviously going to be the diameter. Now just solve the quadratic for that length. You can draw phantom lines across from (1,2) to (5,2) which has distance four and then another line down from (5,4) to (5,2) which has distance 2. 2^2 + 4^2 = 20 = the diameter^2, take the square root of that and you get the answer: \sqrt{20}
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Re: m21 #20 [#permalink]  04 May 2010, 21:01
ans: B
i have not check whether lines are perpendicular or not ,
just assume and got the ans.

but thanks to jallenmorris for giving this concept
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Re: m21 #20 [#permalink]  04 May 2010, 23:20
Hi,

So are we saying if any right angled triangle is in circle its hyptoneuse will always be the diameter of circle whether this statement will always hold true, can you confirm that...?? may sound a bit silly question but havent come across such theorem

Regards,
Sandeep
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Re: m21 #20 [#permalink]  05 May 2010, 07:54
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gmat2104 wrote:
Hi,

So are we saying if any right angled triangle is in circle its hyptoneuse will always be the diameter of circle whether this statement will always hold true, can you confirm that...?? may sound a bit silly question but havent come across such theorem

Regards,
Sandeep

Any two connecting points on the part of a circle is a chord. The chord with the longest distance is
that line that passes through the center. All other chords are minor to this one chord - the diameter.

Extend the two opposite points of the diameter (or the major chord) to meet at any point on the circumference.
Recall that angle at the center is 180 degrees (twice angle that where the two points meet - 90 deg)
In effect, the hypotenuse is the diameter and the geometry formed is a right-angled triangle.

Hope this helps.
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Re: m21 #20 [#permalink]  06 May 2010, 12:31
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my approach ::
ppl might find this weird....

draw an X and Y axis and plot the points roughly..
the points (1,2) and (5,4) clearly forms the diameter, now one would wonder why not point ( 2,5 ) a point on the diameter..and to tht is u have got 3 points given ur expected to get to the ans using this info, (2,5) falls b/w the othr two coordinates..
use coordinates (1,2) and (5,4) as diameter to get the ans
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Re: m21 #20 [#permalink]  27 May 2010, 05:49
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Not that tough question if you knew 2 concepts:
1) Distance between 2 points
2) Length of hypotenuse of right triangle is the diameter is a circle.

Attachment:

dia of circle.png [ 86.14 KiB | Viewed 5490 times ]

A(1,2) , B(2,5) , C(5,4)

Length of AB = \sqrt{(2-1)^2+ (5-2)^2}= \sqrt{10}
Length of BC = \sqrt{(5-2)^2+ (4-5)^2}= \sqrt{10}
Length of CA = \sqrt{(5-1)^2 + (4-2)^2}= \sqrt{20}

From above , we can say that the triangle ABC is isosceles rt triangle.
CA^2=AB^2+BC^2
CA = \sqrt{AB^2+BC^2}= \sqrt{20} . So its a property of a rt. triangle.
Now Hypotenuse of a rt. triangle falls on the diameter of any circle, so dia. is \sqrt{20}
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Re: m21 #20 [#permalink]  06 May 2011, 03:46
The answer is B - root(20). On plotting a graph, it can be seen that this is a right triangle and hypotenuse is root(20).
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Re: m21 #20 [#permalink]  06 May 2011, 05:14
perpendicular lines have inverted slope with sign change..this is the key...
good approach...thanks...i just learnt this concept from MGMAT today and it comes back to me as a revision...thanks SVP
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Re: m21 #20 [#permalink]  09 May 2011, 03:39
The distance between points is sqrt10, sqrt10, sqrt20. This is 45, 45, 90 triangle. By circle properties if any inscribed triangle has 90 degree angle its hypotenuse passes through its center. So diameter is hypotenuse of triangle =sqrt20.
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Re: m21 #20 [#permalink]  09 May 2011, 05:53
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My approach would be:
Let x,y is the center of the circle, then distance between center and any point on the circle is same equal to R.
Given that (1,2), (2,5) and (5,4) lie on circle, we can have following equations:

(x-1)^2 + (y-2)^2 = (x-5)^2 +(y-4)^2
solve it for
2x+y=9

Next take (2,5) and (5,4)
(x-2)^2 + (y-5)^2 = (x-5)^2 +(y-4)^2
and you will get
3x-y=6

Solve for x and y center is (3,3) and distance i.e. Radius is \sqrt{(5-3)^2+(4-3)^2} = \sqrt{5}

So Diameter is \sqrt{20}
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Re: m21 #20 [#permalink]  14 May 2012, 06:09
newdawn wrote:
My approach would be:
Let x,y is the center of the circle, then distance between center and any point on the circle is same equal to R.
Given that (1,2), (2,5) and (5,4) lie on circle, we can have following equations:

(x-1)^2 + (y-2)^2 = (x-5)^2 +(y-4)^2
solve it for
2x+y=9

Next take (2,5) and (5,4)
(x-2)^2 + (y-5)^2 = (x-5)^2 +(y-4)^2
and you will get
3x-y=6

Solve for x and y center is (3,3) and distance i.e. Radius is \sqrt{(5-3)^2+(4-3)^2} = \sqrt{5}

So Diameter is \sqrt{20}

I did the same. However it's way too long and complex calculation, gmat suggests easier solutions to be within 2 minutes limit
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Re: m21 #20 [#permalink]  13 May 2013, 04:16
Expert's post
If a circle passes through points (1, 2), (2, 5), and (5, 4), what is the diameter of the circle?

A. \sqrt{18}
B. \sqrt{20}
C. \sqrt{22}
D. \sqrt{26}
E. \sqrt{30}

Look at the diagram below:
Attachment:

Inscribed triangle.png [ 13 KiB | Viewed 2524 times ]

Calculate the lengths of the sides of triangle ABC:
AB=\sqrt{10};
BC=\sqrt{10};
AC=\sqrt{20}=\sqrt{2}*\sqrt{10};

As we see the ratio of the sides of triangle ABC is 1:1:\sqrt{2}, so ABC is 45°-45°-90° right triangle (in 45°-45°-90° right triangle the sides are always in the ratio 1:1:\sqrt{2}).

So, we have right triangle ABC inscribed in the circle. Now, a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle , so AC=diameter=\sqrt{20}.

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