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Tangents are drawn from the point $(4, 3)$ to the circle ${x^2} + {y^2} = 9$. The area of the triangle formed by them and the line joining their points of contact is
$\frac{{24}}{{25}}$
$\frac{{64}}{{25}}$
$\frac{{192}}{{25}}$
$\frac{{192}}{5}$
Solution
(c) Equation of $AB$ is $4x + 3y = 9$….(i)
{it is chord of contact}
$OQ = \frac{9}{5}$ [perpendicular distance of $AB$ from origin]
$AQ = \sqrt {O{A^2} – O{Q^2}} = \sqrt {9 – \frac{{81}}{{25}}} = \frac{{12}}{5}$
$AB = 2AQ = \frac{{24}}{5}$
$PQ = \frac{{16 + 9 – 9}}{{\sqrt {16 + 9} }} = \frac{{16}}{5}$
Hence the area $ = \frac{1}{2} \times \frac{{24}}{5} \times \frac{{16}}{5} = \frac{{192}}{{25}}$.
Aliter : Required area $ = \frac{a}{{{h^2} + {k^2}}}{({h^2} + {k^2} – {a^2})^{3/2}}$
$ = \frac{3}{{{4^2} + {3^2}}}{({4^2} + {3^2} – 9)^{3/2}} = \frac{{192}}{{25}}$.
