A and B are fixed points. P is a moveable point such that...

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sujoy

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Here is another problem:

Question: A & B are two fixed points and P is a movable point such that PA:pB is constant. Show that the locus of P is a circle.

This problem is confusing isn't it? I really could not make it out.

Regards,
Sujoy
 
\(\displaystyle \L
\begin{array}{l}
A:\left( {h,k} \right),\quad B:\left( {p,q} \right)\quad \& \quad P(x,y) \\
\frac{{PA}}{{PB}} = \frac{{\sqrt {\left( {x - h} \right)^2 + \left( {y - k} \right)^2 } }}{{\sqrt {\left( {x - p} \right)^2 + \left( {y - q} \right)^2 } }} = r \\
\left( {x - h} \right)^2 + \left( {y - k} \right)^2 = r^2 \left[ {\left( {x - p} \right)^2 + \left( {y - q} \right)^2 } \right] \\
\end{array}\)

Note that if r=1 the statement if false.
If r=1 then the locus is a line.
 
Re: there is another problem :

Hello, Sujoy!

A and B are two fixed points; P is a movable point such that PA:pB is constant.
Show that the locus of P is a circle.
Click to expand...
No, it's not "confusing" . . . It's stated very clearly.

However, the algebra required is intimidating . . .
Code: 
            P (x,y)
            o
           *  *
          *     *
         *        *
        *           *
       *              *
    A o - - - - - - - - o B
    (p,q)             (r,s)
Since  PAPB  =  constant\displaystyle \,\frac{PA}{PB}\;=\;constant, we have:  (x−p)2 + (y−q)2(x−r)2 + (y−s)2  =  k\displaystyle \:\frac{\sqrt{(x-p)^2\,+\,(y-q)^2}}{\sqrt{(x-r)^2\,+\,(y-s)^2}}\;=\;k

Then we have:   (x−p)2 + (y−q)2  =  k(x−r)2 + (y−s)2\displaystyle \;\sqrt{(x-p)^2\,+\,(y-q)^2}\;=\;k\sqrt{(x-r)^2\,+\,(y-s)^2}

Square both sides:   (x−p)2 + (y−q)2  =  k2[(x−r)2 + (y−s)2]\displaystyle \;(x-p)^2\,+\,(y-q)^2\;=\;k^2[(x-r)^2\,+\,(y-s)^2]

Expand:   x2 − 2px + p2 + y2 − 2py + q2  =  k2x2 − 2k2rx + k2r2 + k2y2 − 2k2sy + k2s2\displaystyle \;x^2\,-\,2px\,+\,p^2\,+\,y^2\,-\,2py\,+\,q^2\;=\;k^2x^2\,-\,2k^2rx\,+\,k^2r^2\,+\,k^2y^2\,-\,2k^2sy\,+\,k^2s^2

Rearrange:   k2x2 − x2 − 2k2rx + 2px + k2y2 − 2k2sy + 2qy  =  (p2+q2) − k2(r2+s2)\displaystyle \;k^2x^2\,-\,x^2\,-\,2k^2rx\,+\,2px\,+\,k^2y^2\,-\,2k^2sy\,+\,2qy\;=\;(p^2+q^2)\,-\,k^2(r^2+s^2)

Factor:   (k2−1)x2 − 2(k2p−o)x + (k2−1)y2 − 2(k2s−q)y  =  (p2+q2)−k2(r2+s2)\displaystyle \;(k^2-1)x^2\,-\,2(k^2p-o)x\,+\,(k^2-1)y^2\,-\,2(k^2s-q)y\;=\;(p^2+q^2)-k^2(r^2+s^2)

Divide by (k2−1):    x2 − 2(k2r−p)k2−1x + y2 − 2(k2s−q)k2−1y  =  (p2+q2) − k2(r2+s2)k2−1\displaystyle (k^2-1):\;\;x^2\,-\,\frac{2(k^2r-p)}{k^2-1}x\,+\,y^2\,-\,\frac{2(k^2s-q)}{k^2-1}y \;= \;\frac{(p^2+q^2)\,-\,k^2(r^2+s^2)}{k^2-1}

Complete the square:
x2 − 2(k2r−p)k2−1x + [k2r−pk2−1]2 + y2 − 2(k2s−q)k2−1y + [k2s−qk2−1]2  =  (p2+q2) − k2(r2+s2)k2−1 + [k2r−pk2−1]2 + [k2s−qj2−1]2\displaystyle x^2\,-\,\frac{2(k^2r-p)}{k^2-1}x\,+\,\left[\frac{k^2r-p}{k^2-1}\right]^2\,+\,y^2\,-\,\frac{2(k^2s-q)}{k^2-1}y\,+\,\left[\frac{k^2s-q}{k^2-1}\right]^2\;=\;\frac{(p^2+q^2)\,-\,k^2(r^2+s^2)}{k^2-1}\,+\,\left[\frac{k^2r-p}{k^2-1}\right]^2\,+\,\left[\frac{k^2s-q}{j^2-1}\right]^2

Simplify:   (x−k2r−pk2−1)2 + (y − k2s−qk2−1)2  =  k2[(p−r)2 + (q−s)2](k2−1)2\displaystyle \;\left(x - \frac{k^2r-p}{k^2-1}\right)^2\,+\,\left(y\,-\,\frac{k^2s-q}{k^2-1}\right)^2\;=\;\frac{k^2[(p-r)^2\,+\,(q-s)^2]}{(k^2-1)^2}


The locus is a circle with center (k2r−pk2−1, k2s−qk2−1) \displaystyle \left(\frac{k^2r-p}{k^2-1},\,\frac{k^2s-q}{k^2-1}\right)\, and radius  k(p−r)2 + (q−s)2k2−1\displaystyle \,\frac{k\sqrt{(p-r)^2\,+\,(q-s)^2}}{k^2-1}

It is known as the Circle of Apollonius.
 
The solutions are too good & explicit
Thanks a lot .
Regards
Sujoy
 
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