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  • In any circle, the perpendicular from any point on it, to the line joining the points of contact of two tangents is _____of the perpendiculars from the point upon the two tangents ? Fill in the bla

In any circle, the perpendicular from any point on it, to the line joining the points of contact of two tangents is _____of the perpendiculars from the point upon the two tangents ? Fill in the blank.

 

Option: 1

Arithmatic mean


Option: 2

Geometric mean


Option: 3

Harmonic Mean


Option: 4

Square of geometric mean


Answers (1)

best_answer

Let the equation of the circle be x^{2}+y^{2}=a^{2}

Let \mathrm{P\left(a \cos \theta_1, a \sin \theta_1\right)} and \mathrm{Q\left(a \cos \theta_2, a \sin \theta_2\right)} be two points on the circle. Equations of the tangents at P and Q are  

\mathrm{x \cos \theta_1+y \sin \theta_1=a}  _____(1)

and \mathrm{x \cos \theta_2+y \sin \theta_2=a} _____(2)

and the equation of the chord of contact is

\mathrm{x \cos \frac{\left(\theta_1+\theta_2\right)}{2}+y \sin \frac{\left(\theta_1+\theta_2\right)}{2}=a \cos \frac{\left(\theta_1-\theta_2\right)}{2}}

If \mathrm{T}(\mathrm{a} \cos \varphi, \mathrm{a} \sin \varphi) \mathrm{i}

and the equation of the chord of contact is 

\mathrm{"\begin{aligned} & p_1=\left|\frac{a \cos \theta_1 \cos \phi+a \sin \theta_1 \sin \phi-a}{\sqrt{\cos ^2 \theta_1+\sin ^2 \theta_1}}\right|=\left|a\left(\cos \left(\phi-\theta_1\right)-1\right)\right| \\ & p_1=2 a \sin ^2 \frac{\phi-\theta_1}{2} \end{aligned}}

similarly, \mathrm{p_2=2 a \sin ^2 \frac{\phi-\theta_2}{2}}

and \mathrm{p_3=\left|a \cos \phi \cos \frac{\left(\theta_1+\theta_2\right)}{2}+a \sin \phi \sin \frac{\left(\theta_1+\theta_2\right)}{2}-a \cos \frac{\left(\theta_1-\theta_2\right)}{2}\right|}

 

Hence,\mathrm{p_3^2=4 a^2 \sin ^2 \frac{\left(\phi-\theta_1\right)}{2} \sin ^2 \frac{\left(\phi-\theta_2\right)}{2}=p_1 p_2}

\mathrm{\Rightarrow \mathrm{p}_3} is the geometric mean of \mathrm{\Rightarrow \mathrm{p}_1} and \mathrm{\Rightarrow \mathrm{p}_2}.

 

 

 

 

Posted by

manish

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