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  • Let PQ be a diameter of the circle . If <img alt="\alpha" src="htt

Let PQ be a diameter of the circle \mathrm{x^2+y^2=9}. If \alpha and \beta are the lengths of the perpendiculars from P and Q on the straight line, \mathrm{x+y=2} respectively, then the maximum value of \mathrm{\alpha \beta} is __________.

Option: 1

6


Option: 2

7


Option: 3

5


Option: 4

3


Answers (1)

The given equation of circle is  \mathrm{x^2+y^2=3^2 }

\mathrm{ \therefore \text { Radius }=3 }

\mathrm{\text { Let } P(3 \cos \theta, 3 \sin \theta) \text { and } Q(-3 \cos \theta,-3 \sin \theta)}

Given line is \mathrm{x+y=2}

Given \alpha and \beta are the lengths of perpendiculars from P and Q on \mathrm{x+y=2} respectively.

\mathrm{ \begin{aligned} & \therefore \quad \alpha=\frac{|3 \cos \theta+3 \sin \theta-2|}{\sqrt{1+1}} \text { and } \beta=\frac{|-3 \cos \theta-3 \sin \theta-2|}{\sqrt{1+1}} \\\\ & \therefore \quad \alpha \beta=\left|\frac{(3 \cos \theta+3 \sin \theta-2)(3 \cos \theta+3 \sin \theta+2)}{2}\right| \\\\ & \quad=\left|\frac{9+18 \sin \theta \cos \theta-4}{2}\right|=\left|\frac{9(1+\sin 2 \theta)-4}{2}\right| \\\\ & \therefore \quad \text { Maximum value of } \alpha \beta=\frac{9(1+1)-4}{2}=\frac{18-4}{2}=\frac{14}{2}=7 \end{aligned} }

Posted by

Ramraj Saini

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