(iii) Find the value of n for which the circles in the figure have 28 intersections.
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Answer:
This can be done without any trigonometry at all. Let the equations of the circles be
(x−x1)2+(y−y1)2=r21,(1)
(x−x2)2+(y−y2)2=r22.(2)
By subtracting (2) from (1) and then expanding, we in fact obtain a linear equation for x and y; after a little rearranging it becomes
−2x(x1−x2)−2y(y1−y2)=(r21−r22)−(x21−x22)−(y21−y22).
(If the circles intersect, this is the equation of the line that passes through the intersection points.) This equation can be solved for one of x or y; let's suppose y1−y2≠0 so that we can solve for y:
y=−x1−x2y1−y2x+….(3)
Substituting this expression for y into (1) or (2) gives a quadratic equation in only x. Then the x-coordinates of the intersection points are the solutions to this; the y-coordinates can be obtained by plugging the x-coordinates into (3).
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Dec 11 '12 at 7:42
arkeet
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Jul 2 '18 at 19:48
Don Hatch
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When you say by subtracting the two equations? Do you mean from each other? – smac89 Apr 6 '16 at 16:11
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Easy solution is to consider another plane such that the centers are along an axis.
Given the points (x1,y1) and (x2,y2). We focus on the center point of both circles given by
(x1+x22,y1+y22).
The distance between the centers of the circles is given by
R=(x2−x1)2+(y2−y1)2−−−−−−−−−−−−−−−−−−√.
We can consider the following orthogonal vectors
a⃗ =(x2−x1R,y2−y1R),b⃗ =(y2−y1R,−x2−x1R).
a b coordinate plane
In the (a⃗ ,b⃗ ) plane we get the equations
(a+R/2)2+b2=r21,(a−R/2)2+b2=r22.
Whence
a=r21−r222R,b=±r21+r222−(r21−r22)24R2−R24−−−−−−−−−−−−−−−−−−−−−−√.
The intersection points are given by
(x,y)=12(x1+x2,y1+y2)+r21−r222R2(x2−x1,y2−y1)±122r21+r22R2−(r21−r22)2R4−1−−−−−−−−−−−−−−−−−−−−−√(y2−y1,x1−x2),
where R is the distance between the centers of the circles.
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