Tangents are drawn to $$x^2+y^2=16$$ from the point $$P(0,h)$$. These tangents meet the x-axis at $$A$$ and $$B$$. If the area of $$\Delta PAB$$ is minimum, then $$h=?$$
Radius of circle $$x^2+y^2=16$$ is $$4$$ with centre at $$(0,0)$$
Let $$M$$ be $$(x_1,y_1)$$
$$\therefore 2x+2y\cfrac{dy}{dx}=0 \\ \Rightarrow \cfrac{dy}{dx}=\cfrac{-x}{y} \\ \therefore\left| \cfrac{dy}{dx}\right|_{at\;x=x_1,y_1}=\cfrac{-x_1}{y_1}$$
Equation of line PB:
$$(y-y_1)=\cfrac{-x_1}{y_1}(x-x_1)$$
Put $$(0,h)$$\Rightarrow h-y_1=\cfrac{-x_1}{y_1}(0-x_1) \\ \Rightarrow h-y_1=\cfrac{{x_1}^2}{y_1} \Rightarrow h=\cfrac{{x_1}^2}{y_1}+y_1 \\ \Rightarrow h=\cfrac{{x_1}^2+{y_1}^2}{y_1}=\cfrac{16}{y_1} [\because x^2+y^2=16]$$
Put $$y=0$$ in line PB:
$$-y_1=\cfrac{-x_1}{y_1}(x-x_1)\Rightarrow \cfrac{{y_1}^2}{x_1}+x_1=x \\ x=\cfrac{{x_1}^2+{y_1}^2}{x_1}=\cfrac{16}{x_1}$$
$$\therefore$$ Coordinates of $$B\equiv \left( \cfrac { 16 }{ { x }_{ 1 } } ,0 \right) $$
by Similarity,
Coordinates of $$A\equiv \left( \cfrac { 16 }{ { x }_{ 1 } } ,0 \right) $$
$$\therefore \triangle PAB=\cfrac { 1 }{ 2 } \times h\times \left( \cfrac { 32 }{ { x }_{ 1 } } \right) =\cfrac { 16h }{ { x }_{ 1 } } \\ =\cfrac { 16 }{ { x }_{ 1 } } \times \cfrac { 16 }{ { y }_{ 1 } } =\cfrac { { (16) }^{ 2 } }{ { x }_{ 1 }{ y }_{ 1 } } \\ =\cfrac { { (16) }^{ 2 } }{ { x }_{ 1 }\sqrt { 16-{ { x }_{ 1 } }^{ 2 } } } $$
For $$\triangle PAB$$ to be minimum, $${ x }_{ 1 }\sqrt { 16-{ { x }_{ 1 } }^{ 2 } } $$ has to be maximum.
$$g(x)=x\sqrt { 16-{ x }^{ 2 } } \\ \Rightarrow g'(x)=\sqrt { 16-{ x }^{ 2 } } +\cfrac { x(-2x) }{ 2\sqrt { 16-{ x }^{ 2 } } } =0\\ \Rightarrow \sqrt { 16-{ x }^{ 2 } } =\cfrac { { x }^{ 2 } }{ \sqrt { 16-{ x }^{ 2 } } } \Rightarrow 16-{ x }^{ 2 }={ x }^{ 2 }\\ \Rightarrow { x }_{ 1 }=2\sqrt { 2 } \Rightarrow { y }_{ 1 }=\sqrt { 16-{ x }^{ 2 } } =2\sqrt { 2 } \\ \therefore h=\cfrac { 16 }{ { y }_{ 1 } } =\cfrac { 16 }{ 2\sqrt { 2 } } =4\sqrt { 2 } $$
Hence, correct answer is $$4\sqrt{2}$$
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