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From an arbitrary point ‘P’ on the circle \mathrm{x^2+y^2=9}, tangents are drawn to the circle \mathrm{x^2+y^2=1}, which meet \mathrm{x^2+y^2=9} at A and B. Locus of the point of intersection of tangents at A and B to the circle \mathrm{x^2+y^2=9} is

Option: 1

\mathrm{x^2+y^2=\left(\frac{27}{7}\right)^2}


Option: 2

\mathrm{x^2-y^2=\left(\frac{27}{7}\right)^2}


Option: 3

\mathrm{y^2-x^2=\left(\frac{27}{7}\right)^2}


Option: 4

none of these


Answers (1)

best_answer

Since ΔPOQ and ΔAOQ are congruent. Hence ∠POQ = ∠QOA = θ 

\mathrm{\cos \theta=\frac{1}{3} \text {, since } \angle \mathrm{POR}=180^{\circ} \Rightarrow \angle \mathrm{AOR}=\pi-2 \theta}

Now in triangle AOR, ∠AOR = π - 2θ and 

AO = 3 unit

\mathrm{\begin{aligned} & \Rightarrow \cos (\pi-2 \theta)=\frac{O A}{O R}=\frac{3}{\sqrt{h^2+\mathrm{k}^2}} \\ & \Rightarrow \sqrt{\mathrm{h}^2+\mathrm{k}^2}=\frac{27}{7} \Rightarrow \mathrm{x}^2+\mathrm{y}^2=\left(\frac{27}{7}\right)^2 \end{aligned}}

 

Posted by

seema garhwal

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