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The area of the triangle formed by the tangents from the points $(h, k)$ to the circle ${x^2} + {y^2} = {a^2}$ and the line joining their points of contact is
$a{\rm{ }}\frac{{{{({h^2} + {k^2} - {a^2})}^{3/2}}}}{{{h^2} + {k^2}}}$
$a{\rm{ }}\frac{{{{({h^2} + {k^2} - {a^2})}^{1/2}}}}{{{h^2} + {k^2}}}$
$\frac{{{{({h^2} + {k^2} - {a^2})}^{3/2}}}}{{{h^2} + {k^2}}}$
$\frac{{{{({h^2} + {k^2} - {a^2})}^{1/2}}}}{{{h^2} + {k^2}}}$
Solution
(a) Equation of chord of contact $AB $ is
$xh + yk = {a^2}$…..$(i)$
$OM = $length of perpendicular from $O(0, 0)$ on line $(i)$
$ = \frac{{{a^2}}}{{\sqrt {{h^2} + {k^2}} }}$
$\therefore $ $AB = 2AM = 2\sqrt {O{A^2} – O{M^2}} $
$= \frac{{2a\sqrt {{h^2} + {k^2} – {a^2}} }}{{\sqrt {{h^2} + {k^2}} }}$
Also $PM = $length of perpendicular from $P(h,\;k)$to the line $(i)$ is $\frac{{{h^2} + {k^2} – {a^2}}}{{\sqrt {{h^2} + {k^2}} }}$
Therefore, the required area of triangle $PAB$
$ = \frac{1}{2}.\;AB\;.\;PM $
$= \frac{{a{{({h^2} + {k^2} – {a^2})}^{3/2}}}}{{{h^2} + {k^2}}}$.
