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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the locus of the point of intersection of two perpendicular tangents to a given circle $x^2+y^2=a^2$?

$\begin{array}{1 1}(a)\;x^2+y^2=2a^2\\(b)\;x^2+y^2=a^2\\(c)\;x+y=a^2\\(d)\;2x^2+2y^2=3a^2\end{array}$

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1 Answer

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The equation of tangent to the circle $x^2+y^2=a^2$ is
$y=mx+a\sqrt{(1+m^2)}$
$P(h,k)$ lies on tangent then
$(k-mh)=a\sqrt{1+m^2}$
$(k-mh)^2=a^2(1+m^2)$
$m^2(h^2-a^2)-2mhk+k^2-a^2=0$
This is the quadratic equation in m,let two roots $m_1$ and $m_2$
$m_1m_2=-1$($\perp$ tangents)
$\large\frac{k^2-a^2}{h^2-a^2}$$=-1$
$k^2-a^2=-h^2+a^2$
$h^2+k^2=2a^2$
Hence locus P(h,k) is $x^2+y^2=2a^2$
Hence (a) is the correct option.
answered Feb 11, 2014 by sreemathi.v
 

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