A and B are fixed points. P is a moveable point such that...
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\(\displaystyle \L
\begin{array}{l}
A:\left( {h,k} \right),\quad B:\left( {p,q} \right)\quad \& \quad P(x,y) \\
\frac{{PA}}{{PB}} = \frac{{\sqrt {\left( {x - h} \right)^2 + \left( {y - k} \right)^2 } }}{{\sqrt {\left( {x - p} \right)^2 + \left( {y - q} \right)^2 } }} = r \\
\left( {x - h} \right)^2 + \left( {y - k} \right)^2 = r^2 \left[ {\left( {x - p} \right)^2 + \left( {y - q} \right)^2 } \right] \\
\end{array}\)
Note that if r=1 the statement if false.
If r=1 then the locus is a line.
\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!
However, the algebra required is intimidating . . .
Since \(\displaystyle \,\frac{PA}{PB}\;=\;constant\), we have: \(\displaystyle \:\frac{\sqrt{(x-p)^2\,+\,(y-q)^2}}{\sqrt{(x-r)^2\,+\,(y-s)^2}}\;=\;k\)
Then we have: \(\displaystyle \;\sqrt{(x-p)^2\,+\,(y-q)^2}\;=\;k\sqrt{(x-r)^2\,+\,(y-s)^2}\)
Square both sides: \(\displaystyle \;(x-p)^2\,+\,(y-q)^2\;=\;k^2[(x-r)^2\,+\,(y-s)^2]\)
Expand: \(\displaystyle \;x^2\,-\,2px\,+\,p^2\,+\,y^2\,-\,2py\,+\,q^2\;=\;k^2x^2\,-\,2k^2rx\,+\,k^2r^2\,+\,k^2y^2\,-\,2k^2sy\,+\,k^2s^2\)
Rearrange: \(\displaystyle \;k^2x^2\,-\,x^2\,-\,2k^2rx\,+\,2px\,+\,k^2y^2\,-\,2k^2sy\,+\,2qy\;=\;(p^2+q^2)\,-\,k^2(r^2+s^2)\)
Factor: \(\displaystyle \;(k^2-1)x^2\,-\,2(k^2p-o)x\,+\,(k^2-1)y^2\,-\,2(k^2s-q)y\;=\;(p^2+q^2)-k^2(r^2+s^2)\)
Divide by \(\displaystyle (k^2-1):\;\;x^2\,-\,\frac{2(k^2r-p)}{k^2-1}x\,+\,y^2\,-\,\frac{2(k^2s-q)}{k^2-1}y \;= \;\frac{(p^2+q^2)\,-\,k^2(r^2+s^2)}{k^2-1}\)
Complete the square:
\(\displaystyle x^2\,-\,\frac{2(k^2r-p)}{k^2-1}x\,+\,\left[\frac{k^2r-p}{k^2-1}\right]^2\,+\,y^2\,-\,\frac{2(k^2s-q)}{k^2-1}y\,+\,\left[\frac{k^2s-q}{k^2-1}\right]^2\;=\;\frac{(p^2+q^2)\,-\,k^2(r^2+s^2)}{k^2-1}\,+\,\left[\frac{k^2r-p}{k^2-1}\right]^2\,+\,\left[\frac{k^2s-q}{j^2-1}\right]^2\)
Simplify: \(\displaystyle \;\left(x - \frac{k^2r-p}{k^2-1}\right)^2\,+\,\left(y\,-\,\frac{k^2s-q}{k^2-1}\right)^2\;=\;\frac{k^2[(p-r)^2\,+\,(q-s)^2]}{(k^2-1)^2}\)
The locus is a circle with center \(\displaystyle \left(\frac{k^2r-p}{k^2-1},\,\frac{k^2s-q}{k^2-1}\right)\,\) and radius \(\displaystyle \,\frac{k\sqrt{(p-r)^2\,+\,(q-s)^2}}{k^2-1}\)
It is known as the Circle of Apollonius.
Hello, Sujoy!
No, it's not "confusing" . . . It's stated very clearly.A and B are two fixed points; P is a movable point such that PAB is constant.
Show that the locus of P is a circle.
However, the algebra required is intimidating . . .
Code:
P (x,y)
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A o - - - - - - - - o B
(p,q) (r,s)Then we have: \(\displaystyle \;\sqrt{(x-p)^2\,+\,(y-q)^2}\;=\;k\sqrt{(x-r)^2\,+\,(y-s)^2}\)
Square both sides: \(\displaystyle \;(x-p)^2\,+\,(y-q)^2\;=\;k^2[(x-r)^2\,+\,(y-s)^2]\)
Expand: \(\displaystyle \;x^2\,-\,2px\,+\,p^2\,+\,y^2\,-\,2py\,+\,q^2\;=\;k^2x^2\,-\,2k^2rx\,+\,k^2r^2\,+\,k^2y^2\,-\,2k^2sy\,+\,k^2s^2\)
Rearrange: \(\displaystyle \;k^2x^2\,-\,x^2\,-\,2k^2rx\,+\,2px\,+\,k^2y^2\,-\,2k^2sy\,+\,2qy\;=\;(p^2+q^2)\,-\,k^2(r^2+s^2)\)
Factor: \(\displaystyle \;(k^2-1)x^2\,-\,2(k^2p-o)x\,+\,(k^2-1)y^2\,-\,2(k^2s-q)y\;=\;(p^2+q^2)-k^2(r^2+s^2)\)
Divide by \(\displaystyle (k^2-1):\;\;x^2\,-\,\frac{2(k^2r-p)}{k^2-1}x\,+\,y^2\,-\,\frac{2(k^2s-q)}{k^2-1}y \;= \;\frac{(p^2+q^2)\,-\,k^2(r^2+s^2)}{k^2-1}\)
Complete the square:
\(\displaystyle x^2\,-\,\frac{2(k^2r-p)}{k^2-1}x\,+\,\left[\frac{k^2r-p}{k^2-1}\right]^2\,+\,y^2\,-\,\frac{2(k^2s-q)}{k^2-1}y\,+\,\left[\frac{k^2s-q}{k^2-1}\right]^2\;=\;\frac{(p^2+q^2)\,-\,k^2(r^2+s^2)}{k^2-1}\,+\,\left[\frac{k^2r-p}{k^2-1}\right]^2\,+\,\left[\frac{k^2s-q}{j^2-1}\right]^2\)
Simplify: \(\displaystyle \;\left(x - \frac{k^2r-p}{k^2-1}\right)^2\,+\,\left(y\,-\,\frac{k^2s-q}{k^2-1}\right)^2\;=\;\frac{k^2[(p-r)^2\,+\,(q-s)^2]}{(k^2-1)^2}\)
The locus is a circle with center \(\displaystyle \left(\frac{k^2r-p}{k^2-1},\,\frac{k^2s-q}{k^2-1}\right)\,\) and radius \(\displaystyle \,\frac{k\sqrt{(p-r)^2\,+\,(q-s)^2}}{k^2-1}\)
It is known as the Circle of Apollonius.