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In the coordinate plane, a circle has center (2, 3) and [#permalink]
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16 Dec 2009, 04:20
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In the coordinate plane, a circle has center (2, 3) and passes through the point (5, 0). What is the area of the circle? A. 3π B. 3√2π C. 3√3π D. 9π E. 18π
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Last edited by Bunuel on 02 Apr 2012, 01:35, edited 1 time in total.
Edited the question and added the OA



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Re: Coordinate Geometry from Paper test [#permalink]
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16 Dec 2009, 04:24
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IMO E > 18 pi r=sqrt[{52}^2+{0+3}^2]=3 sqrt2.......(dist formula) area=pi r^2 =pi* (3 sqrt2)^2=18 pi
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Re: Coordinate Geometry from Paper test [#permalink]
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16 Dec 2009, 09:51
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E  18pi
r^2 = [(52)^2 + (30)^2] = 18. So Area = pi * r^2 = 18pi



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Re: Coordinate Geometry from Paper test [#permalink]
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17 Dec 2009, 04:55
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OA is E
The distance formula in coordinate geometry is used to calculate the distance between 2 points whose coordinates are given
Lets say we have to calculate the distance between 2 points (x1, y1) and (x2, y2)
It is given by 
sqrt [ (x2x1)^2 + (y2y1)^2]
In this case since we are given the coordinates of the center and the fact that the circle passes thru (5,0), we can calculate the radius (needed for finding the area of the circle), by calculating the distance from the center to point (5,0)
Hope this helps
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Re: Coordinate Geometry from Paper test [#permalink]
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09 Jan 2010, 23:52
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Got E. Used distance formula. Didn't solve for hypotenuse since it was an isoceles right triangle. Plugged in 3 sqrt 2 for radius in area formula. Thanks for the post. Got the answer pretty quickly. What level would you say this is?
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Re: Coordinate Geometry from Paper test [#permalink]
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10 Jan 2010, 00:00
gottabwise wrote: Got E. Used distance formula. Didn't solve for hypotenuse since it was an isoceles right triangle. Plugged in 3 sqrt 2 for radius in area formula. Thanks for the post. Got the answer pretty quickly. What level would you say this is? Just copied problem into notes and recognized how I did extra work...should've just did distance formula b/w (5,0) and (2,3)...r^2=(52)^3+(03)^2=sqrt18=3sqrt2. I guess that's why I'm reviewing right now. Noticing that r^2 doesn't need to be solved for either.
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Re: Coordinate Geometry from Paper test [#permalink]
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02 Apr 2012, 01:01
zaarathelab wrote: In the coordinate plane, a circle has center (2, 3) and passes through the point (5, 0). What is the area of the circle?
A. 3π B. 3√2π C. 3√3π D. 9π E. 18π Can any body draw the picture for this i thought y coordinate between (5,0) & (2,3) will be radius i.e 3 and area will be 3*3pie
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In the coordinate plane, a circle has center (2, 3) and [#permalink]
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02 Apr 2012, 01:34
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GMATD11 wrote: zaarathelab wrote: In the coordinate plane, a circle has center (2, 3) and passes through the point (5, 0). What is the area of the circle?
A. 3π B. 3√2π C. 3√3π D. 9π E. 18π Can any body draw the picture for this i thought y coordinate between (5,0) & (2,3) will be radius i.e 3 and area will be 3*3pie The point is that the radius does not equal to 3, it equals to \(3\sqrt{2}\). You can find the length of the radius either with the distance formula (the formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1x_2)^2+(y_1y_2)^2}\)) or with Pythagoras theorem. Look at the diagram below: The radius of the circle is the hypotenuse of a right isosceles triangle with the legs equal to 3: \(r^2=3^2+3^2=18\) > \(area=\pi{r^2}=18\pi\). Answer: E. Hope it helps. Attachment:
graph.png [ 10.12 KiB  Viewed 38726 times ]
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Re: In the coordinate plane, a circle has center (2, 3) and [#permalink]
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04 Apr 2012, 05:26
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E
we get height as "3" using points (2,0) & (2,3)
we get base as "3" using points (2,0) & (5,0)
therefore (radius)^2 = (3)^2 + (3)^2 radius = 3sqrt2
area = pi * r*r =18pi



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Re: In the coordinate plane, a circle has center (2, 3) and [#permalink]
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The equation of a circle is given by the formula \((xa)^2+(yb)^2=r^2\) where a and b represent the coordinates of the centre and r the radius. Substituting we get \((x2)^2+(y+3)^2=r^2\). Now, given that the circle passes through the points (5,0). Hence, the when we substitute x=5 and y=0 in the equation above it should be satisfied. Substituting we get \((52)^2+(0+3)^2=r^2\). Therefore \(r^2=18\). Area=pi*r^2=18pi.



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Re: Coordinate Geometry from Paper test [#permalink]
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19 Dec 2013, 01:13
Bunuel wrote: GMATD11 wrote: zaarathelab wrote: In the coordinate plane, a circle has center (2, 3) and passes through the point (5, 0). What is the area of the circle?
A. 3π B. 3√2π C. 3√3π D. 9π E. 18π Can any body draw the picture for this i thought y coordinate between (5,0) & (2,3) will be radius i.e 3 and area will be 3*3pie The point is that the radius does not equal to 3, it equals to \(3\sqrt{2}\). You can find the length of the radius either with the distance formula (the formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1x_2)^2+(y_1y_2)^2}\)) or with Pythagoras theorem. Look at the diagram below: Attachment: graph.png The radius of the circle is the hypotenuse of a right isosceles triangle with the legs equal to 3: \(r^2=3^2+3^2=18\) > \(area=\pi{r^2}=18\pi\). Answer: E. Hope it helps. But if the radius = hypothenuse, and if hypothenuse = 18, then the area is 18*18*pi, not 18*pi.. Right? Or am I missing something?



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Re: Coordinate Geometry from Paper test [#permalink]
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19 Dec 2013, 01:17
aeglorre wrote: Bunuel wrote: GMATD11 wrote: Can any body draw the picture for this
i thought y coordinate between (5,0) & (2,3) will be radius i.e 3 and area will be 3*3pie
The point is that the radius does not equal to 3, it equals to \(3\sqrt{2}\). You can find the length of the radius either with the distance formula (the formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1x_2)^2+(y_1y_2)^2}\)) or with Pythagoras theorem. Look at the diagram below: Attachment: graph.png The radius of the circle is the hypotenuse of a right isosceles triangle with the legs equal to 3: \(r^2=3^2+3^2=18\) > \(area=\pi{r^2}=18\pi\). Answer: E. Hope it helps. But if the radius = hypothenuse, and if hypothenuse = 18, then the area is 18*18*pi, not 18*pi.. Right? Or am I missing something? No. Notice that we get that \(r^2=18\), not r. Thus \(area=\pi{r^2}=18\pi\).
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Re: In the coordinate plane, a circle has center (2, 3) and [#permalink]
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25 Jun 2014, 22:30
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I used Ballpark method : (x,y) = center (2,3) to circumference of circle (5,0) that means radius should be more than 3. let's say 3+ area = π*r^2, should be more than π*3^2 ; answer > 9π only 18π is eligible.



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Re: In the coordinate plane, a circle has center (2, 3) and [#permalink]
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01 Feb 2016, 13:07
using the general formula of a circle is the fastest method to solve this problem. (xa)^2 + (yb)^2 = r^2 where (a,b) is the coordinate from which the circle passes through



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Re: In the coordinate plane, a circle has center (2, 3) and [#permalink]
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02 May 2016, 15:32
given: center = (2,3) passes through (5,0) <<< means that a point of the circle, therefore the edge of the circle
let's make a triangle from the center to (5,0)
side 1 = 3, side 2 = 3, radius = x use pythag (we have created a right triangle at the x axis) 3^2+3^2=x^2 9+9=18 3^2+3^2=√18 A=πr^2 A=π(√18)^2 A=18π



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Re: In the coordinate plane, a circle has center (2, 3) and [#permalink]
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03 May 2016, 05:44
zaarathelab wrote: In the coordinate plane, a circle has center (2, 3) and passes through the point (5, 0). What is the area of the circle?
A. 3π B. 3√2π C. 3√3π D. 9π E. 18π Because the circle passes through point (5,0) and has center (2,3), we know that the distance between these points is the radius of the circle. Since we are given the coordinates for each point, the easiest thing to do is to use the distance formula to determine the circle's radius. The distance formula is: Distance= √[(x2 – x1)^2 + (y2 – y1)^2] We are given two ordered pairs, so we can label the following: x1 = 2 x2 = 5 y1 = 3 y2 = 0 When we plug these values into the distance formula, we have: Distance= √[(5 – 2)^2 + (0 – (3))^2] Distance= √ [(3)^2 + (3)^2] Distance= √ [9 + 9] Distance = √ [18] Distance = √9 x √2 Distance = 3 x √2 Thus, we know that the radius = 3 x √2. Finally, we can use the radius to determine the area of the circle. area = ∏r^2 area = ∏(3 x √2 )^2 area = ∏(9 x 2 ) area = 18∏ Answer E.
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Re: In the coordinate plane, a circle has center (2, 3) and [#permalink]
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16 May 2016, 20:44
Attached is a visual that should help.
Attachments
Screen Shot 20160516 at 8.40.06 PM.png [ 143.81 KiB  Viewed 7833 times ]
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Re: In the coordinate plane, a circle has center (2, 3) and [#permalink]
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10 Jul 2016, 20:57
zaarathelab wrote: In the coordinate plane, a circle has center (2, 3) and passes through the point (5, 0). What is the area of the circle?
A. 3π B. 3√2π C. 3√3π D. 9π E. 18π since the circle passes through (5,0) the radius will be the line joining the centre (2,3) and (5,0) Length of radius = \(\sqrt{{3^2+3^2}}\) = \(\sqrt{{18}}\) Area = π*r^2 = π*18 Correct Option: E



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Re: In the coordinate plane, a circle has center (2, 3) and [#permalink]
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25 Nov 2017, 15:04
zaarathelab wrote: In the coordinate plane, a circle has center (2, 3) and passes through the point (5, 0). What is the area of the circle?
A. 3π B. 3√2π C. 3√3π D. 9π E. 18π IMPORTANT: The distance from the center to a point ON the circle = the radius (that's the definition of "radius") What is the distance from (2, 3) to (5, 0)? If we apply the distance formula, we get: Distance = √[(2  5)² + (3  0)²] = √[(3)² + (3)²] = √[9 + 9] = √18 So, the radius is √18Area of circle = π( radius)² = π( √18)² = 18π = E Cheers, Brent
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In the coordinate plane, a circle has center (2, 3) and [#permalink]
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25 Nov 2017, 16:13
zaarathelab wrote: In the coordinate plane, a circle has center (2, 3) and passes through the point (5, 0). What is the area of the circle?
A. 3π B. 3√2π C. 3√3π D. 9π E. 18π To each her own; I find the distance formula cumbersome, especially given a center and a point. I used the general equation for a circle (also called the centerradius form), where \((h,k)\) are the center's coordinates: \((x  h)^2 + (y  k)^2 = r^2\) \(h = 2, k = 3\); this circle's general equation is \((x  2)^2 + (y + 3)^2 = r^2\) Take the point given, (5,0), and plug coordinates into the equation (5 for x, 0 for y): \((5  2)^2 + (0 + 3)^2 = r^2\) \(3^2 + 3^2 = r^2\) \((9 + 9) = 18 = r^2\) Stop there.* Circle's area = \(\pi r^2\), and \(r^2 = 18\) So area = \(\pi r^2 = 18\pi\) Answer E * If you needed to find the radius: \(18 = r^2\) \(\sqrt{r^2}=\sqrt{9*2}\) \(r = 3\sqrt{2}\)
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