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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
    
Associated Topics:
High School Calculus

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