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 positive value of $$h$$ is:

We have the circle $$x^{2}+y^{2}=16$$ whose centre is the origin $$O(0,0)$$ and radius $$r=4$$. From the external point $$P(0,h)$$ two tangents are drawn to the circle. Let one of the tangents meet the x-axis at $$A$$ and the other at $$B$$.

For a circle $$x^{2}+y^{2}=r^{2}$$ the equation of a tangent having slope $$m$$ is stated by the slope form of a tangent:

$$y=mx\;\pm\;r\sqrt{1+m^{2}}.$$

Here $$r=4$$, so every tangent can be written as

$$y=mx\;\pm\;4\sqrt{1+m^{2}}.$$

Because the tangent passes through $$P(0,h)$$ we substitute $$x=0,\;y=h$$ in the above line to obtain the value of its intercept $$c$$:

$$h=\pm 4\sqrt{1+m^{2}}.$$

Squaring both sides we get

$$h^{2}=16(1+m^{2})$$ $$\Rightarrow\;m^{2}=\dfrac{h^{2}}{16}-1.$$

The point $$P$$ lies outside the circle, so $$h>4$$ and hence $$m^{2}>0$$. Therefore the two tangents have slopes

$$m_{1}=+\sqrt{\dfrac{h^{2}}{16}-1}, \qquad m_{2}=-\sqrt{\dfrac{h^{2}}{16}-1}.$$

Because the slopes are equal in magnitude and opposite in sign, the x-intercepts of the tangents are symmetric about the y-axis. Let us calculate the x-intercept of the tangent with positive slope. Putting $$y=0$$ in $$y=mx+h$$ gives

$$0=m\,x+h \;\;\Longrightarrow\;\; x=-\dfrac{h}{m}.$$

Hence the points of intersection with the x-axis are

$$A\bigl(-\dfrac{h}{m},\,0\bigr), \qquad B\bigl(+\dfrac{h}{m},\,0\bigr).$$

Denote $$x_{0}=\dfrac{h}{|m|}=\dfrac{h}{\sqrt{\,h^{2}/16-1\,}}.$$ Thus $$A(-x_{0},0)$$ and $$B(x_{0},0).$$

The triangle $$\triangle APB$$ is isosceles with base $$AB$$ lying on the x-axis and vertex $$P(0,h)$$ lying on the y-axis. Its base length is

$$AB=2x_{0}=2\;\dfrac{h}{\sqrt{\,h^{2}/16-1\,}}.$$

The altitude from $$P$$ to the base is simply $$h$$. Therefore the area $$S(h)$$ of $$\triangle APB$$ is

$$S=\dfrac12 \,(AB)\times h =\dfrac12 \left(2\,\dfrac{h}{\sqrt{\,h^{2}/16-1\,}}\right)h =\dfrac{h^{2}}{\sqrt{\,h^{2}/16-1\,}}.$$

To avoid the complex fraction, multiply numerator and denominator by $$4$$:

$$S(h)=\dfrac{4h^{2}}{\sqrt{h^{2}-16}}, \qquad h>4.$$

We now minimise this function of $$h$$. Differentiate with respect to $$h$$. Write $$S(h)=4h^{2}(h^{2}-16)^{-1/2}$$ and use the product rule $$\dfrac{d}{dh}(uv)=u'\,v+u\,v'$$.

First compute the derivatives: $$\dfrac{d}{dh}(h^{2})=2h,$$ $$\dfrac{d}{dh}\bigl((h^{2}-16)^{-1/2}\bigr)= -\dfrac12 (h^{2}-16)^{-3/2}\cdot 2h =-\dfrac{h}{(h^{2}-16)^{3/2}}.$$

Now,

$$\dfrac{dS}{dh}=4\Bigl[2h\,(h^{2}-16)^{-1/2}+h^{2}\left(-\dfrac{h}{(h^{2}-16)^{3/2}}\right)\Bigr]$$ $$=8h\,(h^{2}-16)^{-1/2}-4h^{3}(h^{2}-16)^{-3/2}.$$

Put everything over the common denominator $$(h^{2}-16)^{3/2}:$$

$$\dfrac{dS}{dh}= \dfrac{8h(h^{2}-16)-4h^{3}}{(h^{2}-16)^{3/2}}.$$

For extrema we set the numerator equal to zero:

$$8h(h^{2}-16)-4h^{3}=0$$ $$\Longrightarrow\;4h\bigl[2(h^{2}-16)-h^{2}\bigr]=0$$ $$\Longrightarrow\;4h\bigl(h^{2}-32\bigr)=0.$$

Since $$h>4$$, we discard the factor $$h=0$$ and obtain

$$h^{2}=32 \;\;\Longrightarrow\;\; h=4\sqrt{2}\;(\text{positive}).$$

To confirm that this point indeed gives a minimum, notice that $$\dfrac{dS}{dh}$$ is negative just above $$h=4$$ (for example at $$h=5$$) and positive for larger $$h$$ (say at $$h=6$$). Hence the function decreases and then increases, so $$h=4\sqrt{2}$$ yields the unique minimum of the area.

Thus the minimum area of $$\triangle APB$$ is attained when

$$h=4\sqrt{2}.$$

Hence, the correct answer is Option 1.