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In the coordinate plane, the points F (2,1), G (1,4), and H
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In the coordinate plane, the points F (2,1), G (1,4), and H (4,1) lie on a circle with center P. What are the coordinates of point P ? A. (0,0) B. (1,1) C. (1,2) D. (1,2) E. (2.5, 2.5)
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Originally posted by gettinit on 21 Nov 2010, 19:47.
Last edited by Bunuel on 10 Jul 2012, 04:29, edited 1 time in total.
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Re: belly of the circle
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21 Nov 2010, 21:23
gettinit wrote: In the coordinate plane, the points F (2,1), G (1,4), and H (4,1) lie on a circle with center P. What are the coordinates of point P ? (A) (0,0) (B) (1,1) (C) (1,2) (D) (1,2) (E) (2.5, (2.5) Remember that center of the circle is equidistant from any point ont he circle. This distance is of course the radius of the circle. The fastest method here is to use the options to see which point is equidistant from F (2,1), G (1,4) and H (4,1). Formula for distance between two points is given by \(\sqrt{(x1  x2)^2 + (y1  y2)^2}\) I see that (0, 0) will not be equidistant from the given 3 points. But distance of (1, 1) from (2, 1) is \(\sqrt{(2  1)^2 + (1  1)^2}\) = 3 Distance of (1, 1) from (1, 4) is \(\sqrt{(1  1)^2 + (1  4)^2}\) = 3 Distance of (1, 1) from (4, 1) is \(\sqrt{(4  1)^2 + (1  1)^2}\) = 3 The coordinates of the center of the circle, P, must be (1, 1). (What I actually did was plotted the points on the xy axis and then saw that (1, 1) will be equidistant from all three points because (2, 1) lies directly 3 steps to its left, (1, 4) lies directly 3 steps above it and (4, 1) lies directly 3 steps to its right.
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Re: belly of the circle
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22 Nov 2010, 11:55
Got it wrong because didn't read the question correctly I thought I need to find the 4th point that would lie on the circle Hence my answer was D (1,2) But the the center of the circle will lie on (1,1)
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Re: belly of the circle
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23 Nov 2010, 19:39
VeritasPrepKarishma wrote: gettinit wrote: In the coordinate plane, the points F (2,1), G (1,4), and H (4,1) lie on a circle with center P. What are the coordinates of point P ? (A) (0,0) (B) (1,1) (C) (1,2) (D) (1,2) (E) (2.5, (2.5) Remember that center of the circle is equidistant from any point ont he circle. This distance is of course the radius of the circle. The fastest method here is to use the options to see which point is equidistant from F (2,1), G (1,4) and H (4,1). Formula for distance between two points is given by \(\sqrt{(x1  x2)^2 + (y1  y2)^2}\) I see that (0, 0) will not be equidistant from the given 3 points. But distance of (1, 1) from (2, 1) is \(\sqrt{(2  1)^2 + (1  1)^2}\) = 3 Distance of (1, 1) from (1, 4) is \(\sqrt{(1  1)^2 + (1  4)^2}\) = 3 Distance of (1, 1) from (4, 1) is \(\sqrt{(4  1)^2 + (1  1)^2}\) = 3 The coordinates of the center of the circle, P, must be (1, 1). (What I actually did was plotted the points on the xy axis and then saw that (1, 1) will be equidistant from all three points because (2, 1) lies directly 3 steps to its left, (1, 4) lies directly 3 steps above it and (4, 1) lies directly 3 steps to its right. Thanks Karishma I also drew it out, and solved that way. But knowing the distance forumla is key. Thank you!



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Re: belly of the circle
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23 Nov 2010, 19:39
amneetpadda wrote: Got it wrong because didn't read the question correctly I thought I need to find the 4th point that would lie on the circle Hence my answer was D (1,2) But the the center of the circle will lie on (1,1) Did you find the alternate point using the distance formula as well? thanks.



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Re: belly of the circle
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23 Nov 2010, 22:05
a very calculation oriented question.... Is there a way of choosing the option to first check, or we should take options in sequential order only...
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Re: belly of the circle
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24 Nov 2010, 06:03
puneetpratik wrote: a very calculation oriented question....
Is there a way of choosing the option to first check, or we should take options in sequential order only... Actually, the only points that will be equidistant from (1, 4) and (4, 1) are (0, 0) and (1, 1) (points where x and y coordinates are same. I will explain why in a minute). When I plot these points on the coordinate axis, I see (0, 0) is way too close to (2, 1). I also see that (1, 1) is perfect because each point is a distance 3 away, either horizontally or vertically. The diagram will show you what I mean: Attachment:
Ques.jpg [ 8.26 KiB  Viewed 14173 times ]
All three colored lines show the distance. Even with a rough sketch, it is apparent. Now, back to 'why only (0, 0) and (1, 1) are possible candidates'. Look at the diagram above. The points (4, 1) and (1, 4) are mirror images of each other (reflected along the line y = x ). Their x and y coordinates are interchanged. These points will be equidistant only from a point lying on y = x i.e. a point whose x and y coordinates are the same. So the calculations actually required in the question are nil.
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Re: belly of the circle
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10 Jul 2012, 04:14
Plug and chug, guys. The answer options contain many small integers that one can easily manipulate in calculations. Alternatively, one can solve it algebraically: Using formula (finding the distance between two points) where x and y are the coordinates of center P. Left side of equation finds the distance between F and P, and right side finds that between G and P. Since both distance are equal (ie. they are both radii), we can equate them thus: (x 2)^2 + (y1)^2 = (x1)^2 + (y1)^2 > x+y=2 Only option B has x and y that satisfy this resultant equation. Cheers, Der alte Fritz.
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Re: belly of the circle
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02 Aug 2012, 10:49
VeritasPrepKarishma wrote: gettinit wrote: In the coordinate plane, the points F (2,1), G (1,4), and H (4,1) lie on a circle with center P. What are the coordinates of point P ? (A) (0,0) (B) (1,1) (C) (1,2) (D) (1,2) (E) (2.5, (2.5) Remember that center of the circle is equidistant from any point ont he circle. This distance is of course the radius of the circle. The fastest method here is to use the options to see which point is equidistant from F (2,1), G (1,4) and H (4,1). Formula for distance between two points is given by \(\sqrt{(x1  x2)^2 + (y1  y2)^2}\) I see that (0, 0) will not be equidistant from the given 3 points. But distance of (1, 1) from (2, 1) is \(\sqrt{(2  1)^2 + (1  1)^2}\) = 3 Distance of (1, 1) from (1, 4) is \(\sqrt{(1  1)^2 + (1  4)^2}\) = 3 Distance of (1, 1) from (4, 1) is \(\sqrt{(4  1)^2 + (1  1)^2}\) = 3 The coordinates of the center of the circle, P, must be (1, 1). (What I actually did was plotted the points on the xy axis and then saw that (1, 1) will be equidistant from all three points because (2, 1) lies directly 3 steps to its left, (1, 4) lies directly 3 steps above it and (4, 1) lies directly 3 steps to its right. Can a solution be figured without plotting the points?



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Re: belly of the circle
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02 Aug 2012, 11:50
navigator123 wrote: Can a solution be figured without plotting the points? Pure algebraic solution: (which I don't recommend but since you asked...) Let the center be (x, y). Distance of each of the three points from the center will be the same. \(\sqrt{(2  x)^2 + (1  y)^2} = \sqrt{(1  x)^2 + (4  y)^2}\) 4x + 4 + 1  2y = 1  2x + 16  8y 6x + 6y  12 = 0 x + y  2 = 0 .......(I) \(\sqrt{(4  x)^2 + (1  y)^2} = \sqrt{(1  x)^2 + (4  y)^2}\) 16  8x + 1  2y = 1  2x + 16  8y 6x  6y = 0 x  y = 0 ........(II) Solving (I) and (II) x = 1, y = 1 Center is (1, 1)
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Re: In the coordinate plane, the points F (2,1), G (1,4), and H
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02 Aug 2012, 15:03
gettinit wrote: In the coordinate plane, the points F (2,1), G (1,4), and H (4,1) lie on a circle with center P. What are the coordinates of point P ?
A. (0,0) B. (1,1) C. (1,2) D. (1,2) E. (2.5, 2.5) FH is parallel to the Xaxis (they both have the same y coordinate). Let's denote by M the middle point of the line segment FH. Then M has coordinates (1, 1) = ((2+4)/2 =1, 1). Because G has its x coordinate 1, the same as that of M, it means that GM is perpendicular to FH (being parallel to the Yaxis) . It follows that the triangle FGH is isosceles, and because FM = MH = GM = 3, it is a right isosceles triangle. Therefore, FH is the diameter of the circumscribed circle, so in fact M is P. Another approach would be to prove that FG is perpendicular to GH (using the slopes). It is quite straightforward to compute them, you will get 1 and 1. Then, immediately, it follows that the triangle is a right triangle, so then P is the middle of FH. Answer B Note: on the GMAT, you will never be asked to find the center of the circle circumscribed to a nonspecial triangle. In the above type of questions, look for the right or maybe equilateral triangle.
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Re: In the coordinate plane, the points F (2,1), G (1,4), and H
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07 Aug 2012, 23:22
The points (2,1) and (4,1) form the diameter of the circle. So the center lies somewhere between these two points. Since y coordinate does not change we need to see where the X coordinate falls. It becomes clear that 1 is the x coordinate. So the point is (1,1). @veritasKarishma Do you think this approach will work always??
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Re: In the coordinate plane, the points F (2,1), G (1,4), and H
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08 Aug 2012, 04:14
rajathpanta wrote: The points (2,1) and (4,1) form the diameter of the circle. So the center lies somewhere between these two points. Since y coordinate does not change we need to see where the X coordinate falls. It becomes clear that 1 is the x coordinate. So the point is (1,1).
@veritasKarishma Do you think this approach will work always?? We know that the three points lie on the same circle but we don't know that (2,1) and (4,1) form the diameter of the circle. The given 3 points could be any 3 points on the circle. So no, given some different values, this approach may not work.
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Re: In the coordinate plane, the points F (2,1), G (1,4), and H
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08 Aug 2012, 06:29
VeritasPrepKarishma wrote: rajathpanta wrote: The points (2,1) and (4,1) form the diameter of the circle. So the center lies somewhere between these two points. Since y coordinate does not change we need to see where the X coordinate falls. It becomes clear that 1 is the x coordinate. So the point is (1,1).
@veritasKarishma Do you think this approach will work always?? We know that the three points lie on the same circle but we don't know that (2,1) and (4,1) form the diameter of the circle. The given 3 points could be any 3 points on the circle. So no, given some different values, this approach may not work. In our case we do know that (2,1) and (4,1) form the diameter of the circumscribed circle, because FG is perpendicular to GH (we can check by computing the slopes of the lines FG and GH, see my previous post). Two points uniquely determine a diameter, but of course, it depends on the third point whether it is on that specific circle. A right triangle is inscribed in a circle having its hypotenuse as diameter (inscribed angle of 90 degree is half of the central angle of 180 degree). What I meant in my previous post was that GMAT will not ask for any nonspecial triangle to find its circumscribed circle's center. It's no point testing distance computations, besides knowing of course the property of the center. For a right triangle, one has to know about testing perpendicularity using slopes, one has to know that a right angle is inscribed in a halfcircle, so the hypotenuse is the diameter...many things to test beside just distance calculation. Similarly, for an equilateral triangle. Regarding areas as well, I cannot remember GMAT asking for the area of some nonspecial triangle. Either one can compute the area by adding/subtracting some areas of some other known geometrical shapes or the triangle is special  right, isosceles, equilateral (well, except the easy case when we have a base and the corresponding height explicitly given). GMAT is not testing advanced geometry formulas for areas like Heron's formula, not even 0.5absinC. That's why I suggest always first test what type of triangle we have.
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Re: In the coordinate plane, the points F (2,1), G (1,4), and H
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08 Aug 2012, 13:34
Theoretically couldn't we know that (2,1) and (4,1) make up the diameter of the circle? Because both points are "on the circle" as the question states, no two points on that axis (where y=1) can be closer or further away, hence it must be the widest part of the circle and the diameter. From there we can solve the midpoint of those two coordinates (making point G unnecessary) so...
(2+4)/2 = 1 for x
(1+1)/2 = 1 for y
so (1,1) must be the center of that line as well as the circle.



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Re: In the coordinate plane, the points F (2,1), G (1,4), and H
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08 Aug 2012, 13:46
Nevermind I see the incredible error in my logic, but once you find their midpoint and see it is equidistant from point G, then you can confirm that it is indeed the midpoint of the entire circle. So point G is not unnecessary as I stated before.



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Re: In the coordinate plane, the points F (2,1), G (1,4), and H
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10 Aug 2012, 11:57
bscharm wrote: Theoretically couldn't we know that (2,1) and (4,1) make up the diameter of the circle? Because both points are "on the circle" as the question states, no two points on that axis (where y=1) can be closer or further away, hence it must be the widest part of the circle and the diameter. From there we can solve the midpoint of those two coordinates (making point G unnecessary) so...
(2+4)/2 = 1 for x
(1+1)/2 = 1 for y
so (1,1) must be the center of that line as well as the circle. Hi, Even I had the same question. Now I am a little confused. Can anyone explain this concept please!!
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Re: In the coordinate plane, the points F (2,1), G (1,4), and H
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Updated on: 10 Aug 2012, 14:03
rajathpanta wrote: bscharm wrote: Theoretically couldn't we know that (2,1) and (4,1) make up the diameter of the circle? Because both points are "on the circle" as the question states, no two points on that axis (where y=1) can be closer or further away, hence it must be the widest part of the circle and the diameter. From there we can solve the midpoint of those two coordinates (making point G unnecessary) so...
(2+4)/2 = 1 for x
(1+1)/2 = 1 for y
so (1,1) must be the center of that line as well as the circle. Hi, Even I had the same question. Now I am a little confused. Can anyone explain this concept please!! You know that FH is the diameter after you check that FG is perpendicular to GH. Every right triangle is inscribed in a circle with the hypotenuse its diameter. Only a right triangle can be inscribed in a half circle. And if one of the sides of a triangle is a diameter in the circumscribed circle, then the triangle must be a right triangle, and the diameter is its hypotenuse. Please, refer to my previous posts above.
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Originally posted by EvaJager on 10 Aug 2012, 13:37.
Last edited by EvaJager on 10 Aug 2012, 14:03, edited 2 times in total.



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Re: In the coordinate plane, the points F (2,1), G (1,4), and H
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10 Aug 2012, 13:50
EvaJager wrote: You know that FH is the diameter after you check that FG is perpendicular to GH. Every right triangle is inscribed in a circle with the hypotenuse its diameter. Please, refer to my previous posts above. I realized this after I had made my initial post, therefore I agree with your post(s). Also your explanation is much better than mine!



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Re: In the coordinate plane, the points F (2,1), G (1,4), and H
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10 Aug 2012, 22:46
rajathpanta wrote: bscharm wrote: Theoretically couldn't we know that (2,1) and (4,1) make up the diameter of the circle? Because both points are "on the circle" as the question states, no two points on that axis (where y=1) can be closer or further away, hence it must be the widest part of the circle and the diameter. From there we can solve the midpoint of those two coordinates (making point G unnecessary) so...
(2+4)/2 = 1 for x
(1+1)/2 = 1 for y
so (1,1) must be the center of that line as well as the circle. Hi, Even I had the same question. Now I am a little confused. Can anyone explain this concept please!! It is certainly correct that both points lie on the circle and no points on the circle on y=1 can be closer or farther away. But what says that these points make a diameter and not just a chord smaller than the diameter? If you do find their mid point, you need to check whether the distance from the third point is equal too. As I said in my first post, you can very easily solve the question by just plotting the points. You can see that all points are equidistant from (1, 1) (Check out a figure I made in a previous post) As for the distance formula, it is a great little tool for any scenario so it's good to know.
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