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In the xy-plane, the point (-2, -3) is the center of a circl [#permalink]
18 Nov 2010, 04:25
3
This post was BOOKMARKED
00:00
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Difficulty:
15% (low)
Question Stats:
75% (02:19) correct
25% (01:17) wrong based on 240 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.
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]
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
1
This post was BOOKMARKED
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.
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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Re: In the xy-plane, the point (-2, -3) is the center of a circl [#permalink]
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).
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