Doug C. answered • 07/18/19
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asked • 07/15/19Derive equation y=kx that cuts through a given circle. Which value is k?
Derive the equation y = kx for a straight line which cuts a 90 degrees arc of the circle
x2 + y2 - 6x - 2y + 5 = 0
The line goes through the origin.
What is the value of k? The answer should have at least 3 decimals.
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Mark M. answered • 07/15/19
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No need for a derivative!
x2 - 6x + 9 + y2 - 2y + 1 = -5 + 9 + 1 complete the square
(x - 3)2 + (y - 1)2 = 5
Now determine the equation of the line passing through the origin and (3 1)
Doug C.
Assuming the line y=kx intersects the circle in two points, the arc intercepted by those two points will be 90 degrees if the lines joining (3,1) to each of those two points are perpendicular. Finding those points of intersection in terms of k looks like it will lead to some tedious results. Wonder if there is another strategy that will work? Going to think about it for a while :).
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07/16/19
Mark H.
I think I've got it---can someone check my answer?
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07/16/19
Mark H. answered • 07/15/19
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Wrong again!!---See the writeup from Doug C.
Doug C.
Hi Mark H. By experimenting on Desmos with the graph of the circle and the graph of various lines where y = kx, along with the fact that the right triangle created by joining (3,1) with the points of intersection will have a hypotenuse = sqrt(10), I think k has a value very close to 1.1282. By using a value of k suggested by your solution (on Desmos) the triangle created does not "look like" a right triangle. In looking at your diagram I am trying to figure out how the sqrt(a^2 + b^2) represents the distance from (3,1) to the point (3,3k)--I am thinking that might be the fallacy in your solution. Still feels like there must be a strategy that makes this problem simpler than what I have been trying.
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07/16/19
Mark H.
Maybe no more time for this just now, but: That distance is simply an assumption about symmetry with the bottom leg which come directly from the a and b offsets Your answer definitely looks right using DESMOS---I think you are right that my assumption about symmetry was wrong---and i agree it should not be this difficult....
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07/16/19
Mark H.
This time I think I've REALLY got it!! (Where have you heard that before? I get 1.1254 --I'm adding the reasoning to my earlier answer--above
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07/16/19
Doug C.
So, I did figure out a solution for this based on the right isosceles triangle that is formed by drawing radii to the points of intersection of the line y=kx and the circle. Getting to the final value for k involved an equation for the altitude of that triangle, finding the point of intersection of the altitude with y = kx, and then using the distance formula between that point of intersection with (3,1). That distance is sqrt(10)/2). I was able to use Newton's method to determine k=1.12771184906. I really think there are two possible values for k, the 2nd being -0.20463. I plan to post this question again and create a video for the solution.
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07/18/19
Mark H.
With any kind of luck, one of us will find a way to reconcile my 1.1254 with your 1.1277....!! I hope the student is learning something from all this......
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07/18/19
Mark H.
Never mind---I see it. I assumed the contact point was at 45 degrees
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07/18/19
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Mark H.
I think that gets you to the center of the circle...when they say "...cuts a 90 degree arc", I interpreted it to mean that the line intercepts the circle in 2 places 90 degrees apart07/15/19