A circle with center O and radius 5 is shown in the xy-plane. Lines that intersect the circle in 2 points include which of the following ?

I. y = -x +1
II. y = 2x + 1
III. y = (1/2)x - 6

A. I only
B. II only
C. I and II only
D. I and III only
E. I, II and III

Find the intercepts by making once x=0 and then y=0. On finding these intercepts, draw line joining these points.
ex for line y=-x+1, the points will be (1,0) and (0,1). On joining these two points, it gets clear that the line will intersect the circle at two points. Similarly for II and III.
You will see that III is far away from the circle.
Hence I and II cut the circle.
+1C
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If a line is intersecting a circle at 2 places, the distance between that line and the center of the circle< The radius

Also, we know that the distance between a point (p,q) and the line ax+by+c=0 is \(\frac{|ap+qb+c|}{\sqrt{a^2+b^2}\)

It's very easy to realise that (p,q) = (0,0) and the distance between the lines in I. and II. are less than 5. For III. the distance =\(\frac{|6|}{\sqrt{\frac{5}{4}}\) = \(\frac{12}{\sqrt{5}}\)>5

Thus, line in option III. doesn't intersect the circle at all.

C
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i did this a much faster way (hopefully it works in all aspects).

I mad x 0 for every equation to find the y intercept.

I. Y = 1
II y = 1
III Y - -6

The circle only spans from -5 to 5 so only I and 2 will intersect in 2 points.

A line that intersects a circle in 2 points is one which has 2 points lying on the circumference of the circle (so it intersects the circumference twice i.e. to say it is not a tangent).
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Why do we assume the center is 0,0? I understand that the radius is 5 so it makes logical sense that the center is 0,0 but couldn't the center easily be for example (0,-2) and then the top point of the circle would be (0,3) and bottom be (0,-7). Radius is still 5 but the circle just sits in a different place in the coordinate plane.

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Ah, I didnt see the picture as I was doing the question on my phone and somehow I don't think the picture loaded. Makes sense now- thanks!

Hi,

I understand the approaches provided by experts.

However, in the real test with scrap paper, how can we draw a perfect circle?

The line in choice iii could have intersected the circle just a tiny bit.

Thanks,

The way Bunuel plotted the points, drew the lines, and got the answer, without having to manually calculate anything was by looking at the y-intercept and the slope:

I. y= -x + 1 ... AKA y = \(\frac{-1}{1}\)x+ 1

Y int = +1
Slope = \(\frac{-1}{1}\)(slope is the number before "x"... in this case it's -1)

1) draw a point at your y-int of (0,1)
2) starting from your Y-intercept, which in this case is (0,1), you have to RISE -1, RUN 1 ... that's your second point .... then RISE -1, RUN 1 again and that's your third point on the line, and so on... then connect the points and you have your line with slope of -1/1... this line intercepts the given circle at 2 points

II. y= 2x + 1

Y int = +1
Slope = \(\frac{+2}{1}\)

1) Draw a point at (0,1)
2) Starting from (0,1) RISE 2, RUN 1 and so on, like you did for number I above.

III. y = \(\frac{1}{2}\)x - 6

Y int = -6
Slope = \(\frac{+1}{2}\)

1) Draw a point at (0,-6)
2) Starting from (0,-6), RISE 1, RUN 2 and so on

Number III is the only one that doesn't intersect the circle in 2 points... so answer is C.

I and II Only
Check if point on line exists inside the circle
Check x = 0
I. y = 1 yes
II. y = 1 yes

III. y = -6
Check the slope
It's 1/2 so line is flatter than slope 1; line has point below circle and slanted downward from circle