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In the coordinate plane, a circle has center (2, -3) and

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In the coordinate plane, a circle has center (2, -3) and [#permalink]

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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
[Reveal] Spoiler: OA

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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IMO E ---> 18 pi

r=sqrt[{5-2}^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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E - 18pi

r^2 = [(5-2)^2 + (-3-0)^2] = 18. So Area = pi * r^2 = 18pi

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Re: Coordinate Geometry from Paper test [#permalink]

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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 [ (x2-x1)^2 + (y2-y1)^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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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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New post 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=(5-2)^3+(0--3)^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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New post 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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Re: Coordinate Geometry from Paper test [#permalink]

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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 \(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)) or with Pythagoras theorem. Look at the diagram below:
Attachment:
graph.png
graph.png [ 10.12 KiB | Viewed 30033 times ]
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.
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Re: In the coordinate plane, a circle has center (2, -3) and [#permalink]

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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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New post 10 Nov 2013, 18:04
The equation of a circle is given by the formula \((x-a)^2+(y-b)^2=r^2\) where a and b represent the coordinates of the centre and r the radius. Substituting we get \((x-2)^2+(y+3)^2=r^2\). 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 \((5-2)^2+(0+3)^2=r^2\). Therefore \(r^2=18\). Area=pi*r^2=18pi.

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Re: Coordinate Geometry from Paper test [#permalink]

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New post 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 \(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)) 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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New post 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 \(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)) 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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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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New post 01 Feb 2016, 13:07
using the general formula of a circle is the fastest method to solve this problem.
(x-a)^2 + (y-b)^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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New post 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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New post 03 May 2016, 05:44
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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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New post 16 May 2016, 20:44
Attached is a visual that should help.
Attachments

Screen Shot 2016-05-16 at 8.40.06 PM.png
Screen Shot 2016-05-16 at 8.40.06 PM.png [ 143.81 KiB | Viewed 739 times ]


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Re: In the coordinate plane, a circle has center (2, -3) and [#permalink]

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New post 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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Re: In the coordinate plane, a circle has center (2, -3) and   [#permalink] 23 Jul 2017, 10:53
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