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In the xy-plane, the point (-2, -3) is the center of a circl [#permalink]
18 Nov 2010, 04:25

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Difficulty:

15% (low)

Question Stats:

75% (02:21) correct
25% (01:18) wrong based on 218 sessions

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]
12 Mar 2013, 03:47

1

This post received KUDOS

Expert's post

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]
15 Mar 2013, 14:14

1

This post received KUDOS

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.

More:"All I wish someone had told me about GMAT beforehand" There are many things you want to know before doing the GMAT exam (how is exam day, what to expect, how to think, to do's...), and you have them in this blog, in a simple way

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]
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]
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]
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]
23 Aug 2014, 04:32

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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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