Conference rusure::math

Center has to lie on x = 4; Center = (4,2) is equidistant from
(2,0), (6,0), and (2,4) but it not tangent. Thus, the y-value of 
the center must be less than 2. y = 1 is too low because the curve 
goes through (2,2) (below (2,4)).

To estimate a solution, there are several ways without calculus.
(I've forgotten all my college calculus, anyway.) Try y = 1.5.
(by eyeball, it looks good). See if it crosses the parabola.
If it does not touch, increase y. If it touches in one point, that's 
the answer. If it crosses in two points, decrease y. 
This should converge quickly.

Sorry I don't have an exact method of solution (probably what was
intended).

Sam (CAD::) Levitin

Let the point of tangency of the circle and the parabola
be (a,a^2).  The center of the circle obviously lies on
the line x=4.  Let the coordinates of the center be (4,k)
Since the point (6,0) is on the circle, the equation of
the circle must be (6-4)^2+(0-k)^2=r^2 -> r^2=k^2+4.

The point (a,a^2) also lies on the circle, so we also know
that (a-4)^2+(a^2-k)^2=k^2+4
     (a^2-8a+16)+(a^4-2ka^2+k^2)=k^2+4
(1)  a^4+a^2-8a+12=2ka^2

The general equation of the circle is (x-4)^2+(y-k)^2=r^2
Differentiating this implicitly and evaluating at (a,a^2)
will give the slope of the tangent at (a,a^2), which must
be the same for both curves.
Differentiating gives  2(x-4)+2(y-k)dy/dx=0
                       dy/dx=-(x-4)/(y-k)
evaluating at (a,a^2) gives dy/dx=(a-4)/(a^2-k)
The slope of the parabola at this point is just 2a,
                     .: 2a = (a-4)/(a^2-k)
                        2a^3-2ak=a-4
                        2ak=2a^3-a+4
(2)                     2ka^2=2a^4-a^2+4a

Equating (1) and (2) gives
         a^4+a^2-8a+12 = 2a^4-a^2+4a
     or  a^4-2a^2+12a-12=0

I don't know of any simple way to solve this, but using the
computer to provide an accurate approximation gives

a=1.0809675685

from which (1) tells us that

k=2.5186851530
r=3.2161739474

The equation is (x-4)^2+(y-2.5186851530)^2=3.2161739474^2

This all reminds me of my old calculus classes (senior year, Newton South,
1970, Mr. Manhard).

One of our exercises involved finding the "osculating circle" in
a parabola.  Apparantly the meaning of "osculate" is "to kiss" !
The osculating circle of a parabola is that circle which fits in
the parabola most snugly, i.e. is most tangent.

The .0 problem here may not be asking for the osculating circle,
but it reminds me of the issue.

/Eric

    
    .1 and .2 state the center must lie on X=4.  Why?
    

    Re .3: technically, 'osculating' means that the SECOND derivatives
    are equal as well. That's not the case here: they have opposite
    sign.
    
    re .4: because x=4 is the only solution of the simultaneous
    equations
    	(x-2)^2+(y-0)^2=r^2
    	(x-6)^2+(y-0)^2=r^2
    determined by the known points (2,0) and (6,0) on the circle.

    
    Re .4
    
    Also, if you look at this geometrically, the perpendicular bisector
    of any chord of a circle passes through the center of the circle.
    The perpendicular bisector of the chord with endpoints (2,0) (6,0) is
    the line x=4.  So the center of the unknown circle must lie along
    the line x=4.

    Re .2
    
    	dy/dx=-(x-4)/(y-k)  -> can someone explain to me where the -
    				sign went from this equation?
    
    
    
    
    						Thanks Mike

    
    Whoops!
    
    You're right, equation (2) is wrong because of the dropped sign,
    it should go like this.
    
    dy/dx = -(x-4)/(y-k) evaluated at (a,a^2) gives
    dy/dx = (a-4)/(k-a^2) = 2a since the slopes are equal at the point
                               of tangency
            a-4 = 2ka-2a^3
            2a^3+a-4 = 2ka
    (2)     2a^4+a^2-4a = 2ka^2
    
    equating (1) and (2) gives
    
    a^4+a^2-8a+12=2a^4+a^2-4a
    a^4+4a-12=0
    
    This equation is a little simpler than the other incorrect one,
    but I still can't find any simple solutions.  Using the computer
    to calculate the solution gives
    
    a = 1.5514798318
    
    from which      
    
    k = 1.6179978229
    r = 2.5725312841
    
    so the equation of the circle is approximately
    
    (x - 4)^2 + (y - 1.618 )^2 = 2.573^2          
    
    Sorry about that!

    I'm a little bothered by the fact that a Calculus II problem ends up 
    with a quartic equation that does not factor into the product of
    two polynomials with rational coefficients.
    
    Speculation #1:  The polynomial  a^4 + 4a   - 12 would be easily
    factorable if by mistake one had a^2 + 4a   - 12
    				  or a^4 + 4a^2 - 12
    and therefore the instructor assigned the problem in the mistaken
    belief that the solution would just pop out.
    
    Speculation #2:  if the equation of the circle is

            (x - 4)^2 + (y - 1.618 )^2 = 2.573^2          
    
    one has to wonder if the y-coordinate is in fact (1 + sqrt(5)) / 2
    a.k.a. phi, the golden ratio, and the radius was maybe (phi + 5).
    
    If #2 were the case then possibly there's some clever way to solve
    the problem and get those results, but I came to the same conclusions
    as prior responses (with a different sign error: a^4 - 4a - 12 (!)).

      John    

I'll ask the student what the instructor said the answer should be.
Yes, it does look too difficult for a Calculus II problem.

I didn't get a clear answer from the student.  He was getting swamped
by the course, and had enough new problems to work on that he didn't
care to resurrect an old one.

    re .*
    
    I also end up with a^4 + 4a - 12 = 0 where a is the x-coordinate of the
    point at which the circle and parabola are tangent. In addition to the
    root near a = 1.55 (as previously noted) there is another root a little
    less than -2. Is this other root also a solution to the problem?
    
    Since quartics can be solved analytically, has anybody tried this? Does
    anyone know the "magic substitution" like the one made for cubics?

        Solving quartics by hand is possible but rather painful. 
        I think it is covered in topic 881.
        
        Dan