Geometry DS

[queenisabella]

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.

The OA is E but why isn't it C?

[pemdas]

because you can always draw another line with the same slope; two lines will have the same slopes and will be parallel and obviously not always cross the circle.

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.

The OA is E but why isn't it C?

[tomada]

Mathematically, remember that a line is given by the equation: y=mx+b

m=slope
b= y-intercept

We are given the slope: m=-1/10

Equation of this line is now y= -x/10 + b

For the x-intercept, y=0.

If you set 0 = -x/10 + b, then x/10 = b.

You can see that, when x=10, b=1, which would be tangent to the top of the circle (0,1).

If x=5, b = 1/2, and the point (0,1/2) is within the circle (intersects)

If x = 20, b = 2, and the point (0,2) is outside the circle (does not intersect)

Thus, depending on the value of 'x' in (x,0), the line could intersect the circle, and the two statements are insufficient even when combined.

[tomada]

Mathematically, remember that a line is given by the equation: y=mx+b

m=slope
b= y-intercept

We are given the slope: m=-1/10

Equation of this line is now y= -x/10 + b

For the x-intercept, y=0.

If you set 0 = -x/10 + b, then x/10 = b.

You can see that, when x=10, b=1, which would be tangent to the top of the circle (0,1).

If x=5, b = 1/2, and the point (0,1/2) is within the circle (intersects)

If x = 20, b = 2, and the point (0,2) is outside the circle (does not intersect)

Thus, depending on the value of 'x' in (x,0), the line could intersect the circle, and the two statements are insufficient even when combined.

[tomada]

Mathematically, remember that a line is given by the equation: y=mx+b

m=slope
b= y-intercept

We are given the slope: m=-1/10

Equation of this line is now y= -x/10 + b

For the x-intercept, y=0.

If you set 0 = -x/10 + b, then x/10 = b.

You can see that, when x=10, b=1, which would be tangent to the top of the circle (0,1).

If x=5, b = 1/2, and the point (0,1/2) is within the circle (intersects)

If x = 20, b = 2, and the point (0,2) is outside the circle (does not intersect)

Thus, depending on the value of 'x' in (x,0), the line could intersect the circle, and the two statements are insufficient even when combined.

[pemdas]

You can see that, when x=10, b=1, which would be tangent to the top of the circle (0,1).

you don't mean tangent as *tangent* right? it's not there as slope=!0 for the line to have tangency point on (0,1)

[tomada]

I meant that the line would pass through (0,1), the uppermost point on the circle.

[pemdas]

I meant that the line would pass through (0,1), the uppermost point on the circle.

and not only that point, obviously not a tangent