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Difficulty: 505-555 Level,    Coordinate Geometry,    Geometry,                   
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Re: Coordinate Geometry from Paper test [#permalink]
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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]
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]
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]
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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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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Attached is a visual that should help.
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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 60908 times ]

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


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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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 √18

Area of circle = π(radius
= π(√18
= 18π
= E

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

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 center-radius 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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Re: In the coordinate plane, a circle has center (2, -3) and [#permalink]
To find the area of the circle, we can use the formula for the area of a circle, which is given by A = πr^2, where r is the radius of the circle.

Given that the circle has a center at (2, -3) and passes through the point (5, 0), we can find the radius by calculating the distance between the center and the given point.

Using the distance formula, the radius (r) can be calculated as follows:

r = √[(x2 - x1)^2 + (y2 - y1)^2] = √[(5 - 2)^2 + (0 - (-3))^2] = √[3^2 + 3^2] = √[18] = 3√2

Now that we have the radius, we can calculate the area of the circle:

A = πr^2 = π(3√2)^2 = π(9 * 2) = π(18)

Hence, the correct answer is E. 18π.
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Re: In the coordinate plane, a circle has center (2, -3) and [#permalink]
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