abhinav11 wrote:
Bunuel 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. This implies that the center of the circle is either in I or IV quadrant AND the x coordinate of the center is greater than the radius of this circle. Still not sufficient. Consider the cases below:
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Not sufficient.
(2) The circle intersects the x-axis at two different positive coordinates. This implies that the distance from the center to the x-axis is less than the radius:
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Not sufficient.
(1)+(2) We know that the center is either in I or IV quadrant. We also know that the x coordinate of the center is greater than the radius of this circle and the distance from the center to the x-axis is less than the radius. This implies that the center is below line y=x, which means that the slope of a line passing origin and the center is less than 1. Sufficient.
Answer: C.
Hi Bunuel,
As always great explanation, Just one question you said as the center is below line y=x , means that slope of a line is less than 1 how do you come to that conclusion??
I am sorry I am weak in Co-ordinate geometry so pardon me if I am missing any basic Formulae for Slope ..
Regards,
Abhinav
Dear Abhinav;
The standard equation of the line is y= mx + c, It is given that the line passes through origin, so coordinate will be (0, 0). To satisfy the condition given i.e. (0,0), you will find the modified equation as y= mx
now the question is asking whether m < 1.
if you assume m=1, you will find a straight line y = x, you will find this line bisects (45 deg) the first quadrant, try to solve the question by drawing, you will understand it much better. in equation y = x, each value of y is equal to x
Now coming to question, for m to be less than 1, if you shift the center of circle below the line y= x, then in new equation of line y=mx, for every value of y, you will have greater corresponding value of x, which indeed is the condition we are looking for y / x < 1.
Similarly, if you shift the center of circle above the line y=x, then in new equation of line y=mx, for every value of y, you will lower corresponding value of x.
If you did not get what I am trying to explain, try to draw this and you will understand.