Let PA and PB be the tangent segments drawn f | vedclass

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(B) The distance $OP = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10$.In the right-angled triangle $	riangle OAP$, $OA = r$ and $AP = \sqrt{OP^2 - r^2} = \s

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$5\sqrt{2}$

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(B) The distance $OP = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10$.In the right-angled triangle $	riangle OAP$, $OA = r$ and $AP = \sqrt{OP^2 - r^2} = \sqrt{100 - r^2}$.The area of $	riangle OAP = \frac{1}{2} 	imes OA 	imes AP = \frac{1}{2} r \sqrt{100 - r^2}$.The area of quadrilateral $PAOB = 2 	imes 	ext{Area}(	riangle OAP) = r \sqrt{100 - r^2}$.Let $f(r) = r \sqrt{100 - r^2}$. To maximize $f(r)$, maximize $f(r)^2 = r^2(100 - r^2) = 100r^2 - r^4$.Let $g(r) = 100r^2 - r^4$. Differentiating with respect to $r$: $g'(r) = 200r - 4r^3$.Setting $g'(r) = 0$, we get $4r(50 - r^2) = 0$, so $r^2 = 50$, which means $r = \sqrt{50} = 5\sqrt{2}$.Thus, the area is maximum when $r = 5\sqrt{2}$.

Let $PA$ and $PB$ be the tangent segments drawn from point $P(6, 8)$ to the circle with the centre at origin $O(0, 0)$. The radius $r$ of the circle for which the area of quadrilateral $PAOB$ is maximum, is...

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