Let PQ be a diameter of the circle x^{2}+y^{2 | vedclass

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(B) Let the coordinates of $P$ be $(3 \cos 	heta, 3 \sin 	heta)$.Since $PQ$ is a diameter,the coordinates of $Q$ are $(-3 \cos 	heta, -3 \sin 	het

Vedclass · Accepted Answer

$7$

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(B) Let the coordinates of $P$ be $(3 \cos 	heta, 3 \sin 	heta)$.Since $PQ$ is a diameter,the coordinates of $Q$ are $(-3 \cos 	heta, -3 \sin 	heta)$.The length of the perpendicular from a point $(x_1, y_1)$ to the line $Ax+By+C=0$ is given by $\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$.For the line $x+y-2=0$,the perpendicular lengths are:$\alpha = \frac{|3 \cos 	heta + 3 \sin 	heta - 2|}{\sqrt{1^2+1^2}} = \frac{|3(\cos 	heta + \sin 	heta) - 2|}{\sqrt{2}}$$\beta = \frac{|-3 \cos 	heta - 3 \sin 	heta - 2|}{\sqrt{1^2+1^2}} = \frac{|-(3(\cos 	heta + \sin 	heta) + 2)|}{\sqrt{2}} = \frac{|3(\cos 	heta + \sin 	heta) + 2|}{\sqrt{2}}$Thus,$\alpha \beta = \frac{|(3(\cos 	heta + \sin 	heta) - 2)(3(\cos 	heta + \sin 	heta) + 2)|}{2} = \frac{|9(\cos 	heta + \sin 	heta)^2 - 4|}{2}$Using $(\cos 	heta + \sin 	heta)^2 = 1 + \sin 2	heta$,we get:$\alpha \beta = \frac{|9(1 + \sin 2	heta) - 4|}{2} = \frac{|9 + 9 \sin 2	heta - 4|}{2} = \frac{|5 + 9 \sin 2	heta|}{2}$Since the maximum value of $\sin 2	heta$ is $1$,the maximum value of $\alpha \beta$ is $\frac{5 + 9(1)}{2} = \frac{14}{2} = 7$.

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

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