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In the xy-plane, the point (-2, -3) is the center of a circl [#permalink]
 18 Nov 2010, 05:25
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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 = A. 6 B. 5 C. 4 D. 3 E. 2
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Re: Problem with question Coordinate Geometry [#permalink]
18 Nov 2010, 05:50
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First of all: 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. Answer: B. For more on this issues check Coordinate Geometry chapter of Math Book: math-coordinate-geometry-87652.htmlHope it helps.
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Re: In the xy-plane, the point (-2, -3) is the center of a circl [#permalink]
12 Mar 2013, 04:47
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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.
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Re: Problem with question Coordinate Geometry [#permalink]
18 Nov 2010, 07:23
Nicely explained. Thanks!
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Re: Problem with question Coordinate Geometry [#permalink]
18 Nov 2010, 09:24
Thanks ...  It helped
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Re: Problem with question Coordinate Geometry [#permalink]
20 Nov 2010, 14:26
Bunuel wrote: First of all: 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. Answer: B. For more on this issues check Coordinate Geometry chapter of Math Book: math-coordinate-geometry-87652.htmlHope it helps. 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 .
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Re: Problem with question Coordinate Geometry [#permalink]
03 Jan 2013, 22:07
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 Answer: B
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Re: In the xy-plane, the point (-2, -3) is the center of a circl [#permalink]
12 Mar 2013, 03:02
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?
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Re: In the xy-plane, the point (-2, -3) is the center of a circl [#permalink]
15 Mar 2013, 15:14
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. Solution: Radius=5 Solution B
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Re: In the xy-plane, the point (-2, -3) is the center of a circl
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15 Mar 2013, 15:14
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