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On the coordinate plane , points P and Q are defined by the

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On the coordinate plane , points P and Q are defined by the [#permalink] New post 14 Aug 2013, 13:45
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On the coordinate plane , points P and Q are defined by the coordinates (-1,0) and (3,3), respectively, and are connected to form a chord of a circle which also lies on the plane. If the area of the circle is (25/4) π , what are the coordinates of the center of the circle ?

(A) (1.5,1)
(B) (2,-5)
(C) (0,0)
(D) (1,1.5)
(E) (2,2)
[Reveal] Spoiler: OA

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Re: On the coordinate plane , points P and Q are defined by the [#permalink] New post 14 Aug 2013, 16:18
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Asifpirlo wrote:
On the coordinate plane , points P and Q are defined by the coordinates (-1,0) and (3,3), respectively, and are connected to form a chord of a circle which also lies on the plane. If the area of the circle is (25/4) π , what are the coordinates of the center of the circle ?

(A) (1.5,1)
(B) (2,-5)
(C) (0,0)
(D) (1,1.5)
(E) (2,2)


π r^2 = 25/4 π

r^2 = 25/4

let (x,y) be the center of circle. Distance to (-1,0)

(x+1)^2 + y^2 = r^2

x^2 + 2x + 1 + y^2 = 25/4 -- Equation 1

Distance from x,y to (3,3)

x^2 -6x-6y+y^2 +18 = 25/4 -- Equation 2

Subtract 2 from 1

8x + 6y = 17.


Only D satisfies the above equation
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Re: On the coordinate plane , points P and Q are defined by the [#permalink] New post 14 Aug 2013, 19:37
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Asifpirlo wrote:
On the coordinate plane , points P and Q are defined by the coordinates (-1,0) and (3,3), respectively, and are connected to form a chord of a circle which also lies on the plane. If the area of the circle is (25/4) π , what are the coordinates of the center of the circle ?

(A) (1.5,1)
(B) (2,-5)
(C) (0,0)
(D) (1,1.5)
(E) (2,2)


D

since area = πr^2 = (25/4) π==>r = \frac{5}{2} ==>diameter =5 ........(1

now see diag Point A(-1,0) AND C (3,3)
Draw perpendicular from C to X-AXIS atB (0,3)
therefore ABC is right triangle with angleB = 90
clearl AB = 4 , BC = 3==>according to pythogoras theoremAC = 5 (diameter of circle)
so chord AC is actually diameter...hence mid point of AC will be centre.
mid point = (avg of x co-ordinates , avg of y co-ordinate)
hence mid point co-ordinate will be (\frac{(3-1)}{2} ,\frac{(3+0)}{2}) = (1,1.5)

hence D
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Re: On the coordinate plane , points P and Q are defined by the [#permalink] New post 14 Aug 2013, 22:25
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An alternative way to approach this problem would be to calculate the distance between the center point and the other two points mentioned in the problem. The distance between center point and other two given points should be the same (will be radius) .
Distance between two points can be calculated as square root of ((x2-x1)^2+(y2-y1)^2))

Option -1: Distance 1 is √(6.25+1 )=√7.25 , Distance 2 is √(2.25+4)=√6.25.Hence Incorrect.
Option-2: Distance1 is √(9+25 )=√34 , Distance 2 is √(1+64)=√65.Hence Incorrect.
Option-3: Distance1 is √(1+0 )=√1 , Distance 2 is √(9+9)=√18.Hence Incorrect.
Option-2: Distance1 is √(4+2.25)=√6.25 , Distance 2 is √(4+2.25)=√6.25. Both are equal and hence Correct.
Option-2: Distance1 is √(9+4 )=√13 , Distance 2 is √(1+1)=√2. Hence Incorrect.
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Re: On the coordinate plane , points P and Q are defined by the [#permalink] New post 15 Aug 2013, 02:16
Asifpirlo wrote:
On the coordinate plane , points P and Q are defined by the coordinates (-1,0) and (3,3), respectively, and are connected to form a chord of a circle which also lies on the plane. If the area of the circle is (25/4) π , what are the coordinates of the center of the circle ?

(A) (1.5,1)
(B) (2,-5)
(C) (0,0)
(D) (1,1.5)
(E) (2,2)


really nice works everyone....................
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Re: On the coordinate plane , points P and Q are defined by the [#permalink] New post 25 Aug 2013, 05:41
I m sorry to ask this question.. I am new to GMAT, you may find this question silly but still I cant able to get this.
We have a point on the circle(x,y)=(-1,0) and we have radius r=5/2.
Equation of circle is : (x-a)^2+(y-b)^2 = r^2. By simply substituting above (x,y) and r we can find (a,b) which is the center.
Y WE ARE GETTING 2 EQUATIONS(as done by maaadhu) SOLVING IT TO GET A FINAL EQUATION AND THEN SUBSTITUTING ANSWER OPTIONS AND SEEING? :oops: :oops:
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Re: On the coordinate plane , points P and Q are defined by the [#permalink] New post 25 Aug 2013, 09:38
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shivananthraj wrote:
I m sorry to ask this question.. I am new to GMAT, you may find this question silly but still I cant able to get this.
We have a point on the circle(x,y)=(-1,0) and we have radius r=5/2.
Equation of circle is : (x-a)^2+(y-b)^2 = r^2. By simply substituting above (x,y) and r we can find (a,b) which is the center.
Y WE ARE GETTING 2 EQUATIONS(as done by maaadhu) SOLVING IT TO GET A FINAL EQUATION AND THEN SUBSTITUTING ANSWER OPTIONS AND SEEING? :oops: :oops:


The approach you are talking about allows to get the answer without actually solving the equations.

BTW, how can you solve (-1-a)^2+(0-b)^2 = (5/2)^2 for a and b?
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Re: On the coordinate plane , points P and Q are defined by the [#permalink] New post 25 Aug 2013, 14:24
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shivananthraj wrote:
I m sorry to ask this question.. I am new to GMAT, you may find this question silly but still I cant able to get this.
We have a point on the circle(x,y)=(-1,0) and we have radius r=5/2.
Equation of circle is : (x-a)^2+(y-b)^2 = r^2. By simply substituting above (x,y) and r we can find (a,b) which is the center.
Y WE ARE GETTING 2 EQUATIONS(as done by maaadhu) SOLVING IT TO GET A FINAL EQUATION AND THEN SUBSTITUTING ANSWER OPTIONS AND SEEING? :oops: :oops:



Yes brother you can back solve it from the last equation you got...
but evaluating the reals roots of the equation will be cumbersome.......

Indeed you can solve any gmat problem without applying some formulas ,And just with few basic understandings.
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Re: On the coordinate plane , points P and Q are defined by the [#permalink] New post 26 Aug 2013, 04:06
Thanks Bunuel and Asif for your replies. I got the gap where i fell.
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Re: On the coordinate plane , points P and Q are defined by the [#permalink] New post 19 Jan 2014, 23:31
Area of the circle
pi*r^2 = \frac{25}{4}pi
=> r=\frac{5}{2}

Distance between the given points is
\sqrt{(3-(-1))^2 + (3-0)^2}

=\sqrt{16+9}

=5

Hence the given points are end points of diameter & mid point of these two points is the center of the circle.

mid-point is
(\frac{{3+(-1)}}{2},\frac{{3+0}}{2})

=(1,1.5)
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Re: On the coordinate plane , points P and Q are defined by the   [#permalink] 19 Jan 2014, 23:31
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