Tangents are drawn from the point $(h, k)$ to the circle $x^{2} + y^{2} = a^{2}$ : prove that the area of the triangle formed by them and the straight line joining their points of the contact is $\frac{a (h^{2} + k^{2})^{3 / 2}}{h^{2} + k^{2}}$ .

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Solution

$x^{2} + y^{2} = a^{2}$

Length of tangent from

$p (h, k)$ to circle is

$A P = \sqrt{h^{2} + k^{2} - a^{2}} = P B$

$O A =$ radius $= a$

By Pythagoras theorem,

$O P = \sqrt{{A P}^{2} + {O A}^{2}}$

$= \sqrt{h^{2} + k^{2}}$

$\Rightarrow cos θ = \frac{A P}{O P} = \frac{P C}{A P}$

$\Rightarrow P C = \frac{{A P}^{2}}{O P} = \frac{h^{2} + k^{2} - a^{2}}{\sqrt{h^{2} + k^{2}}}$

$sin θ = \frac{O A}{O P} = \frac{A C}{A P}$

$\Rightarrow A C = \frac{a \sqrt{h^{2} + k^{2} - a^{2}}}{\sqrt{h^{2} + k^{2}}}$

$A B = 2 A C = \frac{2 a \sqrt{h^{2} + k^{2} - a^{2}}}{\sqrt{h^{2} + k^{2}}}$

$(Δ A P B) = \frac{1}{2} \times P C \times A B$

$= \frac{1}{2} (\frac{h^{2} + k^{2} - a^{2}}{\sqrt{h^{2} + k^{2}}}) \times \frac{2 a \sqrt{h^{2} + k^{2} - a^{2}}}{\sqrt{h^{2} + k^{2}}}$

$= \frac{a {(h^{2} + k^{2} - a^{2})}^{3 / 2}}{h^{2} + k^{2}}$
Hence proved.

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T a n g e n t s a r e f r o m t h e p o i n t (h, k) t o t h e c i r c l e x^{2} + y^{2} = a^{2}; t h e n t h e t h e a r e a o f t h e t r i a n g l e f o r m e d b y t h e s t r a i g h t l i n e j o i n i n g t h e i r p o i n t s o f c o n t a c t i s \frac{a {(h^{2} + k^{2} - a^{2})}^{\frac{3}{2}}}{h^{2} + k^{2}} .

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