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Circle C and line K lie in the xy-plane.

by Guest Tue Sep 23, 2008 9:08 pm

Circle C and line K lie in the xy-plane. If circle C is centered at the origin and has radius 1, does line k intersect circle C?

(1) The x-intercept of line k is greater than 1
(2) The slope of line k is -1/10
:?:
sumit
 
 

is the answer 'E'

by sumit Wed Sep 24, 2008 5:42 am

is the answer E
Guest
 
 

Re: Circle C and line K lie in the xy-plane.

by Guest Thu Sep 25, 2008 9:57 pm

From statement 1, we know that line k could be a line with any slope therefore it can intersect circle c and miss it entirely. NOT sufficient.

From statement 2, knowing the slope of line k still does not tell us if it will intersect circle c. MAYBE. NOT sufficient.

Combining both statements, we know that the x intercept of line k is greater than 1 and that the slope is -1/10. Which means it will have a downward slope (from left to right) and the rise/run is 1 down and 10 right or 1 up and 10 left. Since circle c is on the left of the x intercept it would make sense that for every point of x that the line moves to the left, the y will increase by 1/10 or .1. You can see that, since x>1, line k can intersect circle c and miss circle c entirely. Depending on how far right or left the x intercept is, it is possible. Still we are unsure and do not have enough information. INSUFFICIENT.


Anonymous Wrote:Circle C and line K lie in the xy-plane. If circle C is centered at the origin and has radius 1, does line k intersect circle C?

(1) The x-intercept of line k is greater than 1
(2) The slope of line k is -1/10
:?:
RonPurewal
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Re: Circle C and line K lie in the xy-plane.

by RonPurewal Tue Oct 14, 2008 4:07 am

Anonymous Wrote:From statement 1, we know that line k could be a line with any slope therefore it can intersect circle c and miss it entirely. NOT sufficient.

From statement 2, knowing the slope of line k still does not tell us if it will intersect circle c. MAYBE. NOT sufficient.

Combining both statements, we know that the x intercept of line k is greater than 1 and that the slope is -1/10. Which means it will have a downward slope (from left to right) and the rise/run is 1 down and 10 right or 1 up and 10 left. Since circle c is on the left of the x intercept it would make sense that for every point of x that the line moves to the left, the y will increase by 1/10 or .1. You can see that, since x>1, line k can intersect circle c and miss circle c entirely. Depending on how far right or left the x intercept is, it is possible. Still we are unsure and do not have enough information. INSUFFICIENT.


Anonymous Wrote:Circle C and line K lie in the xy-plane. If circle C is centered at the origin and has radius 1, does line k intersect circle C?

(1) The x-intercept of line k is greater than 1
(2) The slope of line k is -1/10
:?:


well played.

here are 2 takeaways.

TAKEAWAY #1
as with other inequalities, it's instructive to consider the EXTREME possibilities allowed under an inequality (statement 1 is an inequality).

TAKEAWAY #2
you absolutely must be able to visualize the meaning of different slopes.
if you don't know what a slope of -1/10 looks like, then this problem is going to be essentially impregnable.
in general, grinding the algebra, in y = mx + b type form, should be a last resort for problems like these; you should try your best to visualize them first.
remember, the existence of problems that CANNOT be solved by traditional, non-conceptual, "homework" means - or can be solved by such means only with great difficulty - is one of the principal raisons d'être of standardized tests.

taking the two statements together, we can conclude insufficiency very quickly, by considering just the following 2 extreme cases:

(a) putting the x-intercept just barely greater than 1, like, say, 1.001.
in this case, the x-intercept is REALLY close to the circle. a line with slope -1/10 is nearly horizontal, so it clearly intersects the circle.

(b) putting the x-intercept WAY out, like, say, 1,000,000.
in this case, it's clear that there is no way the line is going to hit the circle; it will be at y = 100,000 by the time it gets all the way to the y-axis. (again, there's no reason to think quantitatively about this - just think about putting the x-intercept all the way out there, even with a very gentle slope, and you'll get the picture).

there's really no reason to consider intermediate cases; these extremes will lay bare the essence of the problem.
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by Guest Wed Oct 15, 2008 11:02 pm

Ron-

Please help me to understand this problem. i have been scratching my head on this ques but i figured that i messed up the basic concept. How can i visualize the line with slope-0.1

Please please your help would be greatly appreciated on this.........
Pras
 
 

by Pras Fri Oct 17, 2008 4:13 pm

To visualize the lines based on slopes, it is important to know some basic slope concepts -

Positive Slope ---

A line in the co-ordinate system with Positive slope is one which passes through:
1. Origin, I & III quadrant only.
OR
2. I, III and (II or IV) quadrants.

Negative Slope ---

A line in the co-ordinate system with Negative slope is one which passes through:
1. Origin, II & IV quadrant only.
OR
2. II, IV and (I or III) quadrants.

It is easy to visualize if you draw lines in a sheet of paper using the above information. You shall see -
1. The lines with +ve slope always rise from left to right.
2. The lines with -ve slope always rise from right to left.

How to know if slope is almost horizontal in this case?
Ans: We know that the given slope is -1/10 ie. -.1 which is very close to 0. All the negative slopes close to zero would be almost parallel to x-axis rising from right to left.
RonPurewal
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by RonPurewal Fri Nov 07, 2008 4:34 am

Guest Wrote:Ron-

Please help me to understand this problem. i have been scratching my head on this ques but i figured that i messed up the basic concept. How can i visualize the line with slope-0.1

Please please your help would be greatly appreciated on this.........


hey.

negative slopes go downward as you proceed from left to right. if you imagine compass headings drawn on the coordinate plane (in the traditional way, with north at the top), then negative slopes go from the northwest to the southeast.

if you have a small number (whether small negative or small positive) as your slope, then it's a gentle (almost flat) line. by contrast, slopes greater than 1 are 'steep' lines.

taking these facts together, you have the interpretation of a line with slope -0.1: it slopes down from left to right, but gently.

hope this helps.
acethegmat
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Re: Circle C and line K lie in the xy-plane.

by acethegmat Sun Nov 15, 2009 11:10 am

Ron,

In one of your Manhattan Gmat books it is clearly mentioned:

If slope of line < slope of circle, - no intersection
If slope of line > slope of circle, - 2 intersection
If slope of line = slope of circle, - 1 intersection

Hence, in this question won't St. 2 be enough as we already have the equation of the circle from the question stem?

Please let me know if I have missed something.

Thank you..
esledge
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Re: Circle C and line K lie in the xy-plane.

by esledge Wed Dec 16, 2009 5:02 pm

acethegmat Wrote:Ron,

In one of your Manhattan Gmat books it is clearly mentioned:

If slope of line < slope of circle, - no intersection
If slope of line > slope of circle, - 2 intersection
If slope of line = slope of circle, - 1 intersection

Hence, in this question won't St. 2 be enough as we already have the equation of the circle from the question stem?

Please let me know if I have missed something.

Thank you..

I wasn't able to find where this was mentioned in our strategy guide (post the page # if you have it), but I think there must be some other constraint specified--go back to the book and read it in context.

That other constraint--the one I think you have missed--is the placement of the line relative to the circle, or a specified intersection point through which the slope might vary. I can imagine a line of various slopes hitting 0, 1, or 2 points on a circle (i.e. passing above or below the circle, tangent to the circle, or passing through the circle and out the other side), but I'm basically imagining a line passing through a point near the circle, then "tipping" or "rotating" that line in my mind.

To see what you've missed (or rather, assumed), imagine our line for statement (2) with very slight negative slope, passing close to the origin and intersecting the circle. But then slide the same line way up the y-axis, away from the origin. The line might pass above the circle, and never intersect it.
Emily Sledge
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ManhattanGMAT