coolanujdel wrote: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
Please answer in detail.
mevicks work is great and scrupulous.
To answer this, we need to know the nature of the coordinates of the center of this circle.
(1) This means that no part of the circle lies in the second or third quadrant. In other words, the center is one of these forms: (+, +), (+, -), or (+, 0). Let's plug in for the center. If the center is (3, 2), slope of the mentioned line is 2/3 or less than 1, but, if the center is (2, 3), then slope of the mentioned line is 3/2 or more than 1. Hence insufficient. A and D out, B, C, E to focus on.
(2) This doesn't fix the nature of the coordinates of the center of this circle. The center could be below, above or on the line y = x to get different conclusions about the slope of the mentioned line. Choice B out, our answer is among C and E only.
When the two statements are taken together, it limits us to the right side of the y-axis only. The slope of the mentioned line is 1 if the circle is in first quadrant with the axes tangent to it. But, since the circle has to intersect the x-axis at two different positive coordinates, we need to take it little or more down the x-axis [spoiler]
thus making sure that the slope of the mentioned line is less than 1. That's sufficient, hence C is the correct answer![/spoiler]