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by Anchit 777 » Thu Mar 04, 2010 5:44 am
On a plane there are two points A and B, which are 5 units apart. The no of straight lines at a distance of 2 units from A and 3 units from B are:

A> 1
B> 2
C> 3
D> 4
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by sanju09 » Thu Mar 04, 2010 6:16 am
Anchit 777 wrote:On a plane there are two points A and B, which are 5 units apart. The no of straight lines at a distance of 2 units from A and 3 units from B are:

A> 1
B> 2
C> 3
D> 4
I could see [spoiler]only one line[/spoiler] possible.

[spoiler]A[/spoiler]

[spoiler]Why are you sometimes short of number of choices, mate? Create a dummy fifth if it's not a GMAT query, who cares!![/spoiler]
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by Anchit 777 » Thu Mar 04, 2010 8:20 am
That's what I thought too Sanju, but the answer is 3.
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by kstv » Thu Mar 04, 2010 9:34 am
If we draw 2 circles, from A and B with radius 2 and 3. There are three tangents common to both the circles.
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by sanju09 » Fri Mar 05, 2010 2:25 am
kstv wrote:If we draw 2 circles, from A and B with radius 2 and 3. There are three tangents common to both the circles.
Fantastic!! It's admirable!!!
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by florencejennifer » Tue Mar 09, 2010 3:37 am
Answer Explanation:

By tracing a figure we can identify the no. of such possibilities, and we have at least 5 of such lines.













Correct Answer: D
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by sanju09 » Tue Mar 09, 2010 3:46 am
florencejennifer wrote:Answer Explanation:

By tracing a figure we can identify the no. of such possibilities, and we have at least 5 of such lines.













Correct Answer: D
Please show or explain to us your graphics, it's really very interesting, especially when there's no choice D that reads 5.
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by harshavardhanc » Tue Mar 09, 2010 4:27 am
Anchit 777 wrote:On a plane there are two points A and B, which are 5 units apart. The no of straight lines at a distance of 2 units from A and 3 units from B are:

A> 1
B> 2
C> 3
D> 4
it will be 3 :

at the max , you can have 4 common tangents when the circles do not intersect.

But, in this case the circles, with A & B as centers, themselves would have to be tangents to each other. There by reducing the common tangents by 1.
Regards,
Harsha
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