Let the tangents drawn to the circle x^2 + y^ | vedclass

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(A) Let the circle be $x^2 + y^2 = 4^2$, so the radius $r = 4$.Let the tangent from $P(0, h)$ make an angle $\alpha$ with the $y-$axis. Then $\sin \al

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$4\sqrt{2}$

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(A) Let the circle be $x^2 + y^2 = 4^2$, so the radius $r = 4$.Let the tangent from $P(0, h)$ make an angle $\alpha$ with the $y-$axis. Then $\sin \alpha = \frac{r}{OP} = \frac{4}{h}$.Let $	heta$ be the angle the tangent makes with the $x-$axis. Then $	heta = 90^\circ - \alpha$, so $\cos 	heta = \sin \alpha = \frac{4}{h}$.The $x-$coordinate of $B$ is $OB = \frac{r}{\sin 	heta} = \frac{4}{\cos \alpha} = \frac{4}{\sqrt{1 - (4/h)^2}} = \frac{4h}{\sqrt{h^2 - 16}}$.The area of $\Delta APB$ is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2 \cdot OB) 	imes h = OB 	imes h = \frac{4h^2}{\sqrt{h^2 - 16}}$.Let $f(h) = \frac{16h^4}{h^2 - 16}$. To minimize the area, we minimize $f(h)$.$f'(h) = 16 \left[ \frac{4h^3(h^2 - 16) - h^4(2h)}{(h^2 - 16)^2} ight] = 0 \implies 4h^5 - 64h^3 - 2h^5 = 0 \implies 2h^5 = 64h^3$.Since $h > 4$, $h^2 = 32$, so $h = \sqrt{32} = 4\sqrt{2}$.

Let the tangents drawn to the circle $x^2 + y^2 = 16$ from the point $P(0, h)$ meet the $x-$axis at points $A$ and $B$. If the area of $\Delta APB$ is minimum, then $h$ is equal to

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