A circle is drawn on a coordinate plane

[tstj26]

Hi, looking for some help here. This problem comes from a Manhattan CAT Practice Exam.

A circle is drawn on a coordinate plane. If a line is drawn through the origin and the center of that circle, is the line’s slope less than 1?

(1) No point on the circle has a negative x-coordinate.

(2) The circle intersects the x-axis at two different positive coordinates.

I answered E. The correct answer, however, is C. I am a bit confused as to how you can get this.

When I mentally picture all of the different scenarios that meet these two requirements I can't help but think of incredibly large circles where the circle's center falls into either Quadrant I or IV (as it is defined in Statement 1 that all x coordinates of the circle are positive).

Wouldn't the quadrant that the center of the circle falls into effect the slope of the line? And wouldn't the size of the circle effect the slope of the line?

Thanks

[jlucero]

Hi, looking for some help here. This problem comes from a Manhattan CAT Practice Exam.

A circle is drawn on a coordinate plane. If a line is drawn through the origin and the center of that circle, is the line’s slope less than 1?

(1) No point on the circle has a negative x-coordinate.

(2) The circle intersects the x-axis at two different positive coordinates.

I answered E. The correct answer, however, is C. I am a bit confused as to how you can get this.

When I mentally picture all of the different scenarios that meet these two requirements I can't help but think of incredibly large circles where the circle's center falls into either Quadrant I or IV (as it is defined in Statement 1 that all x coordinates of the circle are positive).

Wouldn't the quadrant that the center of the circle falls into effect the slope of the line? And wouldn't the size of the circle effect the slope of the line?

Thanks

The placement of the center of the circle (but not the size of that circle) would affect the slope of the line, but remember what the question asks: "is the line’s slope less than 1?" You can answer this question without knowing exactly where the circle would be located. Here's a picture of a circle where the center of the circle is connected to the origin, creating a slope of 1.

circle-2.png (7.88 KiB) Viewed 3210 times

The circle can't get any larger without moving into a territory of negative x values. And it would have to move downward so that it has at least two places where it connects to the x-axis. So basically, the circle has to move below the line x=y and any line connecting the origin with the center of the circle will have a slope of less than x=1. Together, the answer to the question is "yes" and the answer to this DS problem is C.

[PrateekC782]

But is it not possible for the center of the circle might lie in the fourth quadrant. Because even than the center of circle will have a positive X-Coordinate. But now the slope will be negative.

[RonPurewal]

But is it not possible for the center of the circle might lie in the fourth quadrant. Because even than the center of circle will have a positive X-Coordinate. But now the slope will be negative.

ok, sure, but that's not the issue.

the issue is that, with both statements together, it's impossible to get a slope of 1 or more.
try drawing out some cases; you'll see what happens.

[RonPurewal]

here's a bit more in the way of explanation (with both statements together):
• let's say you have your two x-intercepts, (a, 0) and (b, 0).
• if the line has a slope of exactly 1, then, by symmetry, there will also be two y-intercepts, (0, a) and (0, b). (don't just visualize the symmetry in your head; actually DRAW the situation, and look at it with your eyes.)
• since there are two y-intercepts, the points between them will lie to the left of the y-axis, violating the other condition.

if you try to make the slope greater than 1, the circle will become even bigger, with an even larger arc to the left of the y-axis.

so, that rules out any effort to draw a circle that gives slope ≥ 1.