P and Q are the ends of a diameter of the cir | vedclass

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(A) Let the coordinates of $P$ be $(a \cos 	heta, a \sin 	heta)$. Since $P$ and $Q$ are ends of a diameter,the coordinates of $Q$ are $(-a \cos 	he

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$a+\frac{1}{\sqrt{2}}$

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(A) Let the coordinates of $P$ be $(a \cos 	heta, a \sin 	heta)$. Since $P$ and $Q$ are ends of a diameter,the coordinates of $Q$ are $(-a \cos 	heta, -a \sin 	heta)$.The line is $x+y-1=0$.The perpendicular distance $s$ from $P(a \cos 	heta, a \sin 	heta)$ is $s = \frac{|a \cos 	heta + a \sin 	heta - 1|}{\sqrt{1^2+1^2}} = \frac{|a(\cos 	heta + \sin 	heta) - 1|}{\sqrt{2}}$.The perpendicular distance $t$ from $Q(-a \cos 	heta, -a \sin 	heta)$ is $t = \frac{|-a \cos 	heta - a \sin 	heta - 1|}{\sqrt{2}} = \frac{|a(\cos 	heta + \sin 	heta) + 1|}{\sqrt{2}}$.Let $u = a(\cos 	heta + \sin 	heta)$. Since $\cos 	heta + \sin 	heta \in [-\sqrt{2}, \sqrt{2}]$,$u \in [-a\sqrt{2}, a\sqrt{2}]$.The product $st = \frac{|u-1|}{\sqrt{2}} \cdot \frac{|u+1|}{\sqrt{2}} = \frac{|u^2-1|}{2}$.Since $a > \frac{1}{\sqrt{2}}$,$a^2 > \frac{1}{2}$,so $a\sqrt{2} > 1$. The maximum value of $u^2$ is $2a^2$.Thus,the maximum of $st$ occurs at $u^2 = 2a^2$,i.e.,$u = \pm a\sqrt{2}$.If $u = a\sqrt{2}$,then $s = \frac{a\sqrt{2}-1}{\sqrt{2}} = a - \frac{1}{\sqrt{2}}$ and $t = \frac{a\sqrt{2}+1}{\sqrt{2}} = a + \frac{1}{\sqrt{2}}$.The greater value is $a + \frac{1}{\sqrt{2}}$.

List-$I$	List-$II$
$A$. The equation of the polar of $(4, 3)$ with respect to $C$	$I$. $y+5=0$
$B$. The equation of the tangent at $(9, -5)$ on $C$	$II$. $x=1$
$C$. The equation of the normal at $(-7, -5)$ on $C$	$III$. $3x+8y=27$
$D$. The equation of the diameter passing through $(1, -5)$ and $(1, 3)$	$IV$. $x=9$

$P$ and $Q$ are the ends of a diameter of the circle $x^2+y^2=a^2$ where $a > \frac{1}{\sqrt{2}}$. $s$ and $t$ are the lengths of the perpendiculars drawn from $P$ and $Q$ onto the line $x+y=1$ respectively. When the product $st$ is maximum,the greater value among $s$ and $t$ is

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