Does line k pass through center of the circle? I) The circle : DS Archive
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# Does line k pass through center of the circle? I) The circle

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Does line k pass through center of the circle? I) The circle [#permalink]

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08 Sep 2006, 01:01
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Does line k pass through center of the circle?

I) The circle intersects with y axis at (0,2) and (0, -2)
II) K intersects with axis at point (-3,0)
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08 Sep 2006, 02:02
Does line k pass through center of the circle?

I) The circle intersects with y axis at (0,2) and (0, -2) ---> No ifo about line k INSUFF
II) K intersects with axis at point (-3,0) ---> No info about circle ---> INSUFF

Together, K can't intercept and be on the same line. Answer is "No" SUFF

(C)
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08 Sep 2006, 05:40
E

just because the circle intersects the y axis at 2 and -2 doesnt mean its center is at (0,0). Its center could be (5,0) or and (x,0) really. So it is possible for the line to intersect the center of the circle, but we cant be sure.
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08 Sep 2006, 08:49
E it is..

zelilmaze wrote:
Does line k pass through center of the circle?

I) The circle intersects with y axis at (0,2) and (0, -2)

II) K intersects with axis at point (-3,0)

Combining the two, assuming the circle is radius 2 from (0,0) the line k could just be x=-3 (vertical line) or it could pass throught the circle.

Hence INSUFF
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08 Sep 2006, 10:11
St1: We know that center of circle must be on the x-axis. But we don't know about line k.: INSUFF

St2: k intersect x-axis at (-3,0). No info about circle.: INSUFF

Together: Considering that center of circle will be on x-axis. Only line that can pass through (-3,0) and point (a,0) is x-axis. But we know that line k is not x-axis.: SUFF
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08 Sep 2006, 12:15
I am sticking with E. The circle could still be

y^2 + (x+3)^ = 13

which would mean that k does pass through the line regardless of it slope.
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08 Sep 2006, 12:23
(E) by a similar approch as mattflow.
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08 Sep 2006, 12:27
I don't think S1 implies that the center is on the x-axis.

The center could be anywhere else with two intersecting points on the Y axis.

Going for E on this...
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09 Sep 2006, 11:09
We don't know where the center of the circle is so E ..
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09 Sep 2006, 17:18
There's an infinite number of circles that may pass thru 0,2 and 0,-2.
The fact that k passes thru -3,0 does not help.

E
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13 Feb 2007, 21:13
Futuristic wrote:
There's an infinite number of circles that may pass thru 0,2 and 0,-2.
The fact that k passes thru -3,0 does not help.

E

The answer E but for a different reason. if the circle intersects the Y axis at 0,-2 and 0,2 then y axis is the chord. Now any line which is perpendicular to it and bisecting it is the diameter (here the X axis). thus the center has to be of the form (a,0). Second statement says that K pass through (intersects X axis) at -3,0 => a = -3 ... hence insufficient. Both put together donot yeild the correct answer.
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13 Feb 2007, 21:30
E

Stat1: with just stat1 we have no info abt line k, so insuff

Stat2: with just stat2 we have no info about the center of the circle, so insuff

even using stat1 we cannot say where the center of the circle is since the radius of the circle can vary, just one point on the line also does not give much more information since the slope of the line can vary, so stat 1& stat 2 doesnt help either...
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13 Feb 2007, 21:45
when you say intersects axis, do you mean x axis?
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14 Feb 2007, 08:02
budugu wrote:
Futuristic wrote:
There's an infinite number of circles that may pass thru 0,2 and 0,-2.
The fact that k passes thru -3,0 does not help.

E

The answer E but for a different reason. if the circle intersects the Y axis at 0,-2 and 0,2 then y axis is the chord. Now any line which is perpendicular to it and bisecting it is the diameter (here the X axis). thus the center has to be of the form (a,0). Second statement says that K pass through (intersects X axis) at -3,0 => a = -3 ... hence insufficient. Both put together donot yeild the correct answer.

He is right. This is why the answer is E.

Circle equation: (x-a)^2 + (y-b)^2 = r^2
Pass (0,2): (0-a)^2 + (2-b)^2 = r^2 .........equa1
Pass (0,-2): (0-a)^2 + (-2 - b)^2 = r^2 ........equa2
eqa1 = eqa2
a^2 + (b-2)^2 = a^2 + (b+2)^2
b^2 - 4b + 4 = b^2 +4b + 4
-4b = 4b
b = 0

Therefore: (x-a)^2 + y^2 = r^2
The origin will be somewhere on (a, 0)

Since k pass (-3,0), K can either pass the origin or not pass the origin.

INSUFF. E
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24 Nov 2009, 19:10
Re: Line k   [#permalink] 24 Nov 2009, 19:10
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