A circle is drawn on a coordinate plane. If a line is drawn

The question asks whether the slope is less than 1. The correct answer is C.

M = Y/X, X is always positive but Y may be Positive or negative, so slope may be +ve or -ve... Hence E

M: stands for line's slope

We are NOT asked whether the slope is negative, we are asked whether the slope is less than 1.

Got it my bad, misinterpreted the question.

Isn't it possible that the circle touches the origin and the radius and the center ( the center lies on the x axis and there is no y co-ordinate for it) of the circle is same ?

If center of the circle is on the x-axis the y-coordinate of the center is 0. In this case the slope of the line will be 0 (since the line would be horizontal), so still less than 1.

Also, the circle cannot pass through the origin, because we are told that "the circle intersects the x-axis at two different [u]positive[/u] coordinates".

Thanks Bunuel.

I considered this option for statement 1 alone. Since this is one of the possibilities, we could also take this into consideration for statement 1 right ?

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.

Hi Bunuel,

Please explain the point " the x coordinate of the center is greater than the radius of this circle", which you have derived from statement 1

Thanks,
Sahil

Can this be proven algebracilly, Second I have not understood your statement "AND the x coordinate of the center is greater than the radius of this circle". and also this statement ". This implies that the distance from the center to the x-axis is less than the radius"
Please help and tell how to approach the problem as there will be no graphs available in the test...

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.

Can this be proven algebraically, Second I have not understood your statement "AND the x coordinate of the center is greater than the radius of this circle". and also this statement ". This implies that the distance from the center to the x-axis is less than the radius"
Please help and tell how to approach the problem….

Is there any other way of doing this question.When I'm considering both together.I am not able to confidently visualize it.

I also have the same doubts. Can someone please explain ?

I will try to explain why (1)+(2) is sufficient
Let take a,b coordinate of the center of the circle
If the line has slope = 1, this means a=b (slope = b/a). The only case when a=b when the circle touches both x and y axis.
To satisfy both statements, the circle must have two x-intercepts, and must either be tangent to the y-axis or not intersect the y-axis at all.
1st Case - In which the circle is tangent to the y-axis—the value of a is equal to the circle’s radius, but the value of b is less than the radius. That is, a > b so the line’s slope b/a must be less than 1.
2nd Case - a is greater than the circle’s radius, while b is still less than that radius. Therefore, the slope b/a is also less than 1 in this case.
The slope has to be less than 1 in all cases satisfying both statements.

I just got it, "slope less than 1" means the same with "gradient of the line is less than that of y=x".
Thanks Bunuel !

Hi Bunuel!

Can you please explain what these highlighted parts mean? How do you come to these conclusions? I am unable to solve this question.

Thanks.

[quote="Bunuel"]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:



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



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.