Line intersecting a circle

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JulianMathHelp

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I'm having trouble putting a simple explanation as to why a line can't interest a circle at 3 points. I know it can't, but I'm not exactly sure how to explain it correctly.

My try was: It can't because, since a line can only pass through a circle at most two points.
 
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Are you looking for a formal proof, or just a convincing explanation? Even in the latter case, it depends on what the person you are convincing understands.

But you might try some version of a proof by contradiction. Suppose three points A, B, C on a line are equidistant from some point X. What can you say about the location of X?
 
JulianMathHelp said:
I'm having trouble putting a simple explanation as to why a line can't interest a circle at 3 points. I know it can't, but I'm not exactly sure how to explain it correctly.

My try was: It can't because, since a line can only pass through a circle at most two points.
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Assume there is such a line. It intersect a circle with the center O in points A, B, C (in this order). Consider the triangle ACO. Since OA and OC are radii of the circle, the triangle is isosceles. Note that B is between A and C and OB length is also equal to the radius . This leads to a contradiction, which I'll leave to you as an exercise.
 
Dr.Peterson said:
Are you looking for a formal proof, or just a convincing explanation? Even in the latter case, it depends on what the person you are convincing understands.

But you might try some version of a proof by contradiction. Suppose three points A, B, C on a line are equidistant from some point X. What can you say about the location of X?
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I’m looking for a convincing explanation.
 
Dr.Peterson said:
Are you looking for a formal proof, or just a convincing explanation? Even in the latter case, it depends on what the person you are convincing understands.

But you might try some version of a proof by contradiction. Suppose three points A, B, C on a line are equidistant from some point X. What can you say about the location of X?
Click to expand...
Not sure about the location of x. Is it in the perpendicular bisector of the line?
 
lev888 said:
Assume there is such a line. It intersect a circle with the center O in points A, B, C (in this order). Consider the triangle ACO. Since OA and OC are radii of the circle, the triangle is isosceles. Note that B is between A and C and OB length is also equal to the radius . This leads to a contradiction, which I'll leave to you as an exercise.
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Is the contradiction that no segment dropped form a vertex of a triangle can be equal length to a side length?
 
No, it is actually wrong. I wasn’t thinking clearly. Is the contradiction that only two points on a line can have the same distance to another point off the line?
 
JulianMathHelp said:
I'm having trouble putting a simple explanation as to why a line can't interest a circle at 3 points. I know it can't, but I'm not exactly sure how to explain it correctly. My try was: It can't because, since a line can only pass through a circle at most two points.
Click to expand...
If you have access to a reasonably complete mathematics library then look for the book ELEMENTARY GEOMETRY FROM AN ADVANCED STANDPOINT by Edwin Moise. In it there is a theorem with proof: The line-circle Theorem, If a line intersects the interior of a circle, then it intersects the circle at exactly two points.
 
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Sorry for the messy handwriting, and the crams, but does this proof work?
 

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Let [math](x - h)^2 + (y - k)^2 = r^2[/math] be the circle in question and let [math]y = mx + b[/math] be a line that intersects the circle.

Then
[math](x - h)^2 + (mx + b - k)^2 = r^2[/math]
[math](m^2 + 1) x^2 + (-2h + 2m(b - k)) x + ( h^2 + (b - k)^2 - r^2) = 0[/math]is a quadratic equation. If the line intersects a circle, then there is either one or two real solutions for x (depending if the discriminant is zero or non-zero), which gives us one or two points of intersection.

I leave the details to you.

-Dan
 
topsquark said:
Let [math](x - h)^2 + (y - k)^2 = r^2[/math] be the circle in question and let [math]y = mx + b[/math] be a line that intersects the circle.

Then
[math](x - h)^2 + (mx + b - k)^2 = r^2[/math]
[math](m^2 + 1) x^2 + (-2h + 2m(b - k)) x + ( h^2 + (b - k)^2 - r^2) = 0[/math]is a quadratic equation. If the line intersects a circle, then there is either one solution or two real solutions for x (depending if the discriminant is zero or non-zero), which gives us one or two points of intersection.

I leave the details to you.

-Dan
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Can't you do it without the constants in the circle equation, as it doesn't affect the exponents of the variables?
 
I circle does not have three points that are co-linear. That is the main point.

Draw a circle. Take a straight edge. Put the straight edge horizontal above the circle. Carefully move the straight edge down until the straight edge is below the circle. The whole time notice that the straight edge never crossed three points at once. That would convince me that the statement is true.
 
JulianMathHelp said:
Not sure about the location of x. Is it in the perpendicular bisector of the line?
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Close. A line doesn't have a perpendicular bisector; a segment does.

So X, the center of our circle, has to be on not one, but three different perpendicular bisectors -- which are all parallel to one another! That's not going to happen ...
 
Dr.Peterson said:
Close. A line doesn't have a perpendicular bisector; a segment does.

So X, the center of our circle, has to be on not one, but three different perpendicular bisectors -- which are all parallel to one another! That's not going to happen ...
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Would my proof work though? Also I can’t really understand the last part. Why does it have to be in 3?
 
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JulianMathHelp said:
Would my proof work though? Also I can’t really understand the last part. Why does it have to be in 3?
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Post #11? I haven't tried to read it.

As for mine, don't AB, BC, and AC all have perpendicular bisectors that are parallel? But you only need to use two of them in a proof.
 
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