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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the length of common chord of the two circles $(x-p)^2+(y-q)^2=c^2$ and $(x-q)^2+(y-p)^2=c^2$.

$\begin{array}{1 1}(a)\;\sqrt{4c^2-2(p-q)^2}\\(b)\;\sqrt{4c^2+2(p-q)^2}\\(c)\;\sqrt{4c^2-2(p+q)^2}\\(d)\;\sqrt{6c^2-(p-q)^2}\end{array}$

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1 Answer

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  • Length of $\perp$ from a point $(x_1,y_1)$ on the line $ax+by+c=0$ is $\bigg|\large\frac{ax_1+by_1+c}{\sqrt {a^2+b^2}}\bigg|$
$S_1=(x-p)^2+(y-q)^2-c^2=0$
$S_2=(x-q)^2+(y-p)^2-c^2=0$
Equation of common chord is $S_1-S_2=0$
$\Rightarrow\:(x-p)^2+(y-q)^2-(x-q)^2-(y-p)^2=0$
$\Rightarrow\:-2xp-2yq+2xq+2yp=0$
$\Rightarrow\:2p(y-x)+2q(x-y)=0$
$\Rightarrow\:(p-q)(y-x)=0$
$i.e., $ Equation of common chord $PQ\:is\:y=x$
$C_1M$ is the length of $\perp$ from $C_1(p,q)$ on the line $PQ x-y=0$
$\Rightarrow\:$Length $C_1M$ is equal to $\large\frac{|p-q|}{\sqrt 2}$
$\Rightarrow\:C_1P$=radius of the first circle=$c$
In $\Delta PC_1M$,PM=$\sqrt{(PC_1)^2+(C_1M)^2}$
$\Rightarrow \sqrt{c^2-\large\frac{(p-q)^2}{2}}$
Also $PQ=2PM=2\sqrt{c^2-\large\frac{(p-q)^2}{2}}$
$\Rightarrow \sqrt{4c^2-2(p-q)^2}$
Hence (a) is the correct answer.
answered Feb 12, 2014 by sreemathi.v
edited Mar 20, 2014 by rvidyagovindarajan_1
 

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