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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the locus of a middle point of a system of parallel chords of a circle $x^2+y^2=a^2$?

$\begin{array}{1 1}(a)\;x+my=0\\(b)\;y+mx=0\\(c)\;x^2+my^2=0\\(d)\;\text{None of these}\end{array}$

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1 Answer

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Circle is $x^2+y^2=a^2$ and let equation of parallel chord is $y=mx+c$
P(h,k) be the middle point of chord $y=mx+c$
P is the mid point of $(x_1,y_1)$ and $(x_2,y_2)$ then
$h=\large\frac{x_1+x_2}{2}$$,k=\large\frac{y_1+y_2}{2}$
$x_1+x_2=2h$ and $y_1+y_2=2k$-----(1)
P(h,k) lie on line $y=mx+c$
$k-mh=c$------(2)
Substituting $y=mx+c$ in $x^2+y^2=a^2$
$x^2+(mx+c)^2=a^2$
$(1+m^2)x^2+2mcx+c^2-a^2=0$
If $x_1,x_2$ are roots
$x_1+x_2=-\large\frac{2mc}{1+m^2}$
$2h=-\large\frac{2m(k-mh)}{(1+m^2)}$)[from (1) and (2)]
$h+m^2h=-mk+m^2h$
$\Rightarrow h+mk=0$
$\Rightarrow x+my=0$
Hence (a) is the correct answer.
answered Feb 12, 2014 by sreemathi.v
 

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