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In the xy-plane, the point (-2, -3) is the center of a circl [#permalink]

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18 Nov 2010, 04:25

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In the xy-plane, the point (-2, -3) is the center of a circle. The point (-2, 1) lies inside the circle and the point (4, -3) lies outside the circle. If the radius r of the circle is an integer, then r =

No posting of PS/DS questions is allowed in the main Math forum.

rite2deepti wrote:

In the xy-plane, the point (-2, -3) is the center of a circle. The point (-2, 1) lies inside the circle and the point (4, -3) lies outside the circle. If the radius r of the circle is an integer, then r =

A. 6 B. 5 C. 4 D. 3 E. 2

OA=B

The easiest way to solve this question will be just to mark the points on the coordinate plane. You'll see that the distance between the center (-2, -3) and the point inside the circle (-2, 1) is 4 units (both points are on x=-2 line so the distance will simply be 1-(-3)=4) so the radius must be more than 4, and the distance between the center (-2, -3) and the point outside the circle (4, -3) is 6 units (both points are on y=-3 line so the distance will simply be 4-(-2)=6) so the radius must be less then 6 --> 4<r<6, thus as r is an integer then r=5.

Re: In the xy-plane, the point (-2, -3) is the center of a circl [#permalink]

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12 Mar 2013, 03:47

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mahendru1992 wrote:

Okay i used another method but i'm not getting the answer so I used the equation (x-a)^2+(y-b)^2=r^2, and put in the values of the center (-2,-3), which then comes out to be (x+2)^2+(y+3)^2=r^2. Then i substitute the values of the inside point i.e (-2,1) for x and y and my radius comes out to be 4. I don't know what am i doing wrong here?

What you are doing wrong is that the point (-2,1) doesn't lie ON the circle. You can not substitute this value for the circle's equation.
_________________

Re: In the xy-plane, the point (-2, -3) is the center of a circl [#permalink]

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15 Mar 2013, 14:14

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See the attached drawing.

We can easily see by drawing the problem that a Radius=4 is not enough, as the circle will have its frontier in (-2,1) - see the clear circle - and therefore the point will NOT be inside the circle.

If we increase the radius to the next integer Radius=5, this is solved, as the point (-2,1) will fall inside the circle, while the point (4,-3) will fall outside the circle - see the dark circle -. And this is the only feasible solution to the problem.

No posting of PS/DS questions is allowed in the main Math forum.

rite2deepti wrote:

In the xy-plane, the point (-2, -3) is the center of a circle. The point (-2, 1) lies inside the circle and the point (4, -3) lies outside the circle. If the radius r of the circle is an integer, then r =

A. 6 B. 5 C. 4 D. 3 E. 2

OA=B

The easiest way to solve this question will be just to mark the points on the coordinate plane. You'll see that the distance between the center (-2, -3) and the point inside the circle (-2, 1) is 4 units (both points are on x=-2 line so the distance will simply be 1-(-3)=4) so the radius must be more than 4, and the distance between the center (-2, -3) and the point outside the circle (4, -3) is 6 units (both points are on y=-3 line so the distance will simply be 4-(-2)=6) so the radius must be less then 6 --> 4<r<6, thus as r is an integer then r=5.

What would be more time saving is just to use the distance formula to get the dist between center and point inside the circle as 4 & Distance between the center and outside point as 6 so 4<r<6 as above and Answer is B .
_________________

Re: Problem with question Coordinate Geometry [#permalink]

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03 Jan 2013, 21:07

rite2deepti wrote:

In the xy-plane, the point (-2, -3) is the center of a circle. The point (-2, 1) lies inside the circle and the point (4, -3) lies outside the circle. If the radius r of the circle is an integer, then r =

A. 6 B. 5 C. 4 D. 3 E. 2

OA=B

Simply plot the coordinates and you will figure out that r is greater than 4 but less than 6: 4 < r < 6

Re: In the xy-plane, the point (-2, -3) is the center of a circl [#permalink]

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12 Mar 2013, 02:02

Okay i used another method but i'm not getting the answer so I used the equation (x-a)^2+(y-b)^2=r^2, and put in the values of the center (-2,-3), which then comes out to be (x+2)^2+(y+3)^2=r^2. Then i substitute the values of the inside point i.e (-2,1) for x and y and my radius comes out to be 4. I don't know what am i doing wrong here?

Re: In the xy-plane, the point (-2, -3) is the center of a circl [#permalink]

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01 Aug 2013, 06:36

Could someone help me how go about this? Since (4,-3) lies outside the circle, it is clear that one of the point that lies on the circle is (x,-3). The other point (-2,1) lies inside the circle, so another point on the circle would be (-2,y). We also know that the center is (-2,-3). Join the center to the point (x,-3) and (-2,y) to form the 2 radius. After equating the 2 lines I get (y+3)^2=(x+2)^2. What should I do ahead?

Re: In the xy-plane, the point (-2, -3) is the center of a circl [#permalink]

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23 Aug 2014, 04:32

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: In the xy-plane, the point (-2, -3) is the center of a circl [#permalink]

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30 Jan 2016, 06:48

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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