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 contact is $\frac{a (h^{2} + k^{2} - a^{2})^{\frac{3}{2}}}{h^{2} + k^{2}}$ .

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Solution

Let tangents are drawn from $P (h, k)$ to circle $x^{2} + y^{2} = a^{2}$ at point of contact of tangents be $Q$ and $R$

Equation of chord of contact $Q R$ is

$h x + k y = a^{2} h x + k y - a^{2} = 0...... (i)$

Draw $P S$ perpendicular to $Q R$

$P S = \frac{h^{2} + k^{2} - a^{2}}{\sqrt{h^{2} + k^{2}}}$

length of tangent $= P R = \sqrt{S_{1}}$

$\Rightarrow P R = \sqrt{h^{2} + k^{2} - a^{2}}$

In $△ P S R$ $,$ ${P S}^{2} + {S R}^{2} = {P R}^{2}$

$\Rightarrow {S R}^{2} = {P R}^{2} - {P S}^{2} \Rightarrow {S R}^{2} = {(\sqrt{h^{2} + k^{2} - a^{2}})}^{2} - {(\frac{h^{2} + k^{2} - a^{2}}{\sqrt{h^{2} + k^{2}}})}^{2} \Rightarrow {S R}^{2} = (h^{2} + k^{2} - a^{2}) {1 - \frac{h^{2} + k^{2} - a^{2}}{h^{2} + k^{2}}} \Rightarrow {S R}^{2} = (h^{2} + k^{2} - a^{2}) {\frac{h^{2} + k^{2} - h^{2} - k^{2} + a^{2}}{h^{2} + k^{2}}} \Rightarrow {S R}^{2} = \frac{a^{2} (h^{2} + k^{2} - a^{2})}{h^{2} + k^{2}} \Rightarrow S R = \frac{a \sqrt{h^{2} + k^{2} - a^{2}}}{\sqrt{h^{2} + k^{2}}}$
But $Q R = 2 S R$
$\Rightarrow Q R = \frac{2 a \sqrt{h^{2} + k^{2} - a^{2}}}{\sqrt{h^{2} + k^{2}}}$

Area of $△ P Q R$ $= \frac{1}{2} \times Q R \times P S$

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

$\Rightarrow Δ = \frac{{a (h^{2} + k^{2} - a^{2})}^{\frac{3}{2}}}{h^{2} + k^{2}}$

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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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