How do I prove that the following point lies on director circle? Tangents are drawn to $x^2+y^2=16$ from the point P $(0,h)$. These tangent stuff meet the x-axis at $A$ And $B$. Find h if area of Triangle $PAB$ is minimum.
The answer of h is $sqrt.(32)$ which shows that this point lies on director circle ⭕️. How do I get a proof that intuitively proves this?
If I assume x-intercept of both tangents to be $(a,0) and (-a,0)$ Then area of Triangle is $ah$. Which is in the form of a rectangular hyperbola where I am guessing the minimum value of $ah$ will be when a=h
 A: A calculation of $h$ below:
Look at $\triangle OBP$.(Symmetry about $y-$axis)
Call point of tangency $T$.
$|OT| =r$.
Consider right $\triangle OTP:$
$|PT|^2 = h^2 -r^2 , |PT| =\sqrt{h^2-r^2}.$
For right $\triangle OBP:$
$|PT||TB| = r^2.$
Area of $\triangle OBP:$
Area $= (1/2)(|PT| +|TB|)r=$
$(1/2)(|PT| + r^2/(|PT|)) .$
AM-GM:
$(|PT| +r^2/|PT|) \ge $
$2 \sqrt{(|PT|)(r^2/|PT|)} = 2r$.
Hence : Area $\ge r $.
Equality for $|PT| = r^2/|PT|$, or 
$|PT|= r= 4.$
$h^2= |PT|^2+r^2= 2r^2 = 32$
$h=\sqrt{32}.$
Hence?
A: A bit of trigonometry:
Note the symmetry about $y-$axis:
Look at $\triangle OBP$ (right half of the original).
Call point of tangency $T$.
$\angle BOT = \alpha.$
Then : 
$\tan \alpha =  |BT| /r$
Area $\triangle OBT =(1/2)|BT|r=$
$(1/2)r^2\tan \alpha$. 
Likewise :
Area $\triangle OTP = (1/2)|TP|r=$
$(1/2)r^2\tan (π/2-\alpha)$.
$S:=$ Area $\triangle OBP =$
$(1/2)r^2(\tan \alpha + \tan (π/2-\alpha)).$
Need to minimize $S$:
$S:= \tan \alpha + \tan(π/2- \alpha)$, $0 < \alpha <π/2$.
$z: =\tan \alpha, z>0$.
$S= z +1/z$.
AM GM:
$S \ge 2\sqrt{z(1/z)} =2$.
Equality for $z=1/z$, i.e $z=1$.
Hence
$ \tan \alpha =1$, i.e. $\alpha =π/4$, and
$\angle OPT =π/4.$
A: You have a semicircle above the $x$-axis defined by:
$$f(x):=\sqrt{16-x^2}$$

Let $O=(0,0)$ be the center of the semicircle $f$. To draw a line tangent to $f$, connect find the midpoint of $OP$, which is $M_{OP}=\left(0,\frac h2\right)$. Now, draw a circle centered at $M_{OP}$ with $r=h/2$. Now we defined the function:
$$g(x):=\frac{1}{2} \left(\sqrt{h^2-4 x^2}+h\right)$$
Now the points where $f(x)=g(x)$ are the points of tangency, for which we have:
$$\left(-\frac{4 \sqrt{h^2-16}}{h},\frac{16}{h}\right),\left(\frac{4 \sqrt{h^2-16}}{h},\frac{16}{h}\right)$$

Now, write the functions for the lines tangent to $f$:
$$h_1(x):=\frac{1}{4} \sqrt{h^2-16} x+h\\
h_2(x):=-\frac{1}{4} \sqrt{h^2-16} x+h$$
Now $h_1$ and $h_2$ intersect the $x$-axis (i.e. their zeroes) at:
$$h_1:\left\{x\to -\frac{4 h}{\sqrt{h^2-16}}\right\}\\
h_2:\left\{x\to \frac{4 h}{\sqrt{h^2-16}}\right\}$$
Solving for the distance between these two points (i.e. the length of the base of the triangle), we get:
$$AB=\frac{8 h}{\sqrt{h^2-16}}$$
The height of $\triangle PAB$ is $h$, therefore the area of $\triangle PAB$ is:
$$A_{\triangle PAB}=\frac12\cdot AB\cdot h=\frac12 \cdot \frac{8 h}{\sqrt{h^2-16}} \cdot h = \frac{4 h^2}{\sqrt{h^2-16}}$$

To minimize $A_{\triangle PAB}$, simply take the derivative and set it to $0$:
$$\left(\frac{4 h^2}{\sqrt{h^2-16}}\right)'=0\\
\frac{4 h \left(h^2-32\right)}{\left(h^2-16\right)^{3/2}}=0\\
4 h ( h^2-32 )=0$$
Therefore, you have the following solutions:
$$\left\{\{h\to 0\},\left\{h\to -4 \sqrt{2}\right\},\left\{h\to 4 \sqrt{2}\right\}\right\}$$

We know that $\angle A=\angle B$ because $PAB$ is isosceles. Then angle $A$ is $\arctan m_1$, where $m_1$ is the slope of the line passing through $P$ and $A$. Therefore:
$$\angle A=\arctan \left(\frac{\sqrt{h^2-16}}{4}\right)$$
Since we have $h=\pm4\sqrt2$ as a solution, then:
$$\angle A=\arctan-1=\frac\pi4 \iff \bbox[10px, border:2px black solid]{\therefore P=\frac\pi2}$$
