The area of the triangle formed by the tangents form the point $(4, 3)$ to the circle $x^{2} + y^{2} = 9$ and the line joining their point of contact is:

\frac{192}{27}

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\frac{192}{25}

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\frac{192}{29}

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\frac{192}{31}

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Solution

The correct option is B $\frac{192}{25}$
Since $P A = P B = \sqrt{4^{2} + 3^{2} - 9} = \sqrt{16 + 9 - 9} = 4$

$∠ C P A = ∠ C P B = α$

In $△ P A C, tan α = \frac{3}{4}, sin α = \frac{3}{5}$

$cos α = \frac{P M}{4} = \frac{4}{5}$

$P M = \frac{16}{5}$

$sin α = \frac{A M}{4} = \frac{3}{5}$

$\Rightarrow A M = \frac{12}{5}$

Area of $△ P A B = \frac{1}{2} A B . P M$

$\Rightarrow \frac{1}{2} (2 A M) . P M = A M . P M$

$= \frac{12}{5} \times \frac{16}{5} = \frac{192}{25}$

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Similar questions

(h, k)

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

P (3, 4)

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

A

B

△ P A B = \frac{192}{25}

a =

(h, k)

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

\frac{a (h^{2} + k^{2})^{3 / 2}}{h^{2} + k^{2}}

x^{2} + y^{2} = 25

\frac{192}{- 168}

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