Find the Value of a...Date: 9 Jan 1995 07:13:11 -0500 From: Secret Someone Subject: (none) Here's my problem, Doc: Find the value of "a" so that the circles with equations (x-a)^2+y^2=2 and (x+a)^2+y^2=2 intersect at points where their tangents are perpendicular. So, I just do not get this. I can't even visualize what my math teacher is talking about. I am in BC Calculus so use whatever method you like. It would be great if someone could "shed a little light" on this. Thanks, Steven Burris Date: 9 Jan 1995 14:01:28 -0500 From: Dr. Ken Subject: Re: your mail Hello there! I'm afraid I won't be a huge amount of help on the visualization thing, since you can't really draw pictures on this screen. But I'll give it a shot. Okay, try these things for me: draw two circles that are about the same size, and draw them so that they just barely intersect (not tangent, but so that the two intersection points are pretty close together. Now look at one of the points where they intersect, and look at the "angle" formed by the pieces of the two circles where they meet. Hopefully, you've drawn the circles so that they meet at a pretty sharp angle. Now draw some circles that overlap a little more. The angle (actually two pairs of vertical angles; I'm thinking of the smaller angle) formed at the intersection points should be a little bigger now. And so on. The way we would measure these angles is by drawing the tangent lines to each of the circles at the intersection point, and then looking at the angle between the two tangent lines. In this problem, we're looking for two circles that intersect in a perpendicular way. That said (for what it's worth), we can start to tackle this problem. First of all, recall that the the slope of the tangent line to a function is given by the derivative with respect to x. And also recall that when two lines are perpendicular, their slopes are the opposite reciprocals of each other (like x and -1/x). So let's start. First off, let's find the points of the circles' intersection. Now, don't go off solving complicated systems of equations on this one. Look at what the circle equations are: two circles, each of radius Sqrt{2}, which have their centers at (a,0) and (-a,0). So the intersection points are going to be halfway between them in the x direction, right? I.e. they'll be on the y-axis. So just plug in x=0 to these equations to find out what the two intersection points are. It looks like we get y= +- Sqrt{2-a^2}. So the coordinates of the two points of intersection are (0,Sqrt{2-a^2}) and (0,-Sqrt{2-a^2}). Note that the two intersection points will give exactly the same angles of intersection, so we'll just use one of these points, and we'll pick it according to which one makes the algebra easier. Okay, now we need to differentiate these bad boys. With respect to x. Probably implicit differentiation is going to be easiest here. So we get 2(x-a) + 2y * dy/dx = 0 and 2(x+a) + 2y * dy/dx = 0. So dy/dx = (a-x)/y in the first circle, and dy/dx = (-a-x)/y in the second circle. So now we plug in the coordinates of the intersection point: we'll use the first intersection point to get dy/dx = a/Sqrt{2-a^2} and dy/dx = -a/Sqrt{2-a^2}. THESE ARE THE SLOPES OF THE TANGENT LINES AT THE INTERSECTION POINTS!! So what do we want to do with them again? We want to see which values of a make these two things opposite reciprocals of each other. So we'll set it up. a/Sqrt{2-a^2} = Sqrt{2-a^2}/a So now you solve for a. Sound good? -Ken "Dr." Math Note from Dr. Ken: Several months later, Ujjwal Rane (palvisai@imap1.asu.edu) was browsing through our archives and found this problem, and discovered that there's an easier solution than the one I gave. Here's Ujjwal's solution: - Let c = distance between the centers = 2*a - Since the tangents are perpendicular, their radii must also meet perpendicularly, at the common points. - From the equations of the circles, radius r = Sqrt{2} - From Pythagoras thorem: r^2 + r^2 = c^2 - Now plug and chug: 2 + 2 = c^2 (since r = Sqrt{2}) 4 = c^2 2 = c 2 = 2a => a = 1 Thanks Ujjwal! -Ken "Dr." Math |
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