Prove that a line cannot intersect a circle at more than two point
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Answers
Step-by-step explanation:
The cases for 0, 1, and 2 intersections, respectively, no intersection, tangent, and secant, are all fairly well defined. So the question is: Can a line intersect a circle at more than 2 points?
We prove by contradiction. Suppose we have a circle with three or more collinear points. We then rotate the circle such that those points lie parallel to the x-axis, thus determining the y-value of our intersecting (horizontal) line which we can write as:
y =n
Now we look at the equation of a circle:
(x -h)² - (y-k)² = r²
From our analysis, we already know the values of h, y, k,and r, or rather they are constants as opposed to variables. Taking all non-x to be constant, this leaves us with the general equation:
a x² + bx + c = 0
By the fundamental theorem of algebra, there are exactly 2 roots and by extension at most 2 distinct roots by virtue of the polynomial having an order of 2. However, if our line intersected the circle at more than 2 points then the solution set must contain at least 3 distinct roots (otherwise the overlap point would be redundant) which would violate the fundamental theorem of algebra. Thus, we have arrived at a contradiction.
Therefore, a line cannot intersect a circle at more than two points.
Answer:
The cases for 0, 1, and 2 intersections, respectively, no intersection, tangent, and secant, are all fairly well defined. So the question is: Can a line intersect a circle at more than 2 points?
We prove by contradiction. Suppose we have a circle with three or more collinear points. We then rotate the circle such that those points lie parallel to the x-axis, thus determining the y-value of our intersecting (horizontal) line which we can write as:
y =n
Now we look at the equation of a circle:
( x − h) 2 + y − k 2 = r 2
From our analysis, we already know the values of h, y, k,and r, or rather they are constants as opposed to variables. Taking all non-x to be constant, this leaves us with the general equation:
From our analysis, we already know the values of h, y, k,and r, or rather they are constants as opposed to variables. Taking all non-x to be constant, this leaves us with the general equation:
a x 2 + b x + c = 0
By the fundamental theorem of algebra, there are exactly 2 roots and by extension at most 2 distinct roots by virtue of the polynomial having an order of 2. However, if our line intersected the circle at more than 2 points then the solution set must contain at least 3 distinct roots (otherwise the overlap point would be redundant) which would violate the fundamental theorem of algebra. Thus, we have arrived at a contradiction.
Therefore, a line cannot intersect a circle at more than two points.
Step-by-step explanation:
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