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Re: Coordinate Geometry from Paper test [#permalink]

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17 Dec 2009, 04:55

1

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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)

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? _________________

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=(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. _________________

Re: Coordinate Geometry from Paper test [#permalink]

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02 Apr 2012, 01:34

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Expert's post

3

This post was BOOKMARKED

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 [ 10.12 KiB | Viewed 17031 times ]

The radius of the circle is the hypotenuse of a right isosceles triangle with the legs equal to 3: \(r^2=3^2+3^2=18\) --> \(area=\pi{r^2}=18\pi\).

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

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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.

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 \(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)) 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?

Re: Coordinate Geometry from Paper test [#permalink]

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19 Dec 2013, 01:17

Expert's post

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 \(area=\pi{r^2}=18\pi\). _________________

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

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25 Jun 2014, 22:30

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.

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

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22 Jul 2015, 06:16

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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. (x-a)^2 + (y-b)^2 = r^2 where (a,b) is the coordinate from which the circle passes through

gmatclubot

Re: In the coordinate plane, a circle has center (2, -3) and
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01 Feb 2016, 13:07

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