From a point $ P(0,b) $ two tangents are drawn to the circle $ x^2+y^2=16 $ and these two tangents intersect x-axis at two points A and B .If the area of triangle PAB is minimum ,then prove that the equation of its circumcircle is $ x^2+y^2=32 $.
The solution is given in my book .They wrote area of triangle PAB is minimum if angle PAB is 90 degree . I didn't unterstand the reason . Can anyone give a hint ?
Thanks in advance.
The area of $PAO$ is $\frac{1}{2}\cdot 4\cdot \frac{4}{\sin{\theta}\cos{\theta}}=\frac{16}{\sin{2\theta}}$. Area is minimum if $\sin{2\theta}$ is maximum.
Let $r=4$, $\angle OPA=\phi$, then
\begin{align} b&=\frac r{\sin\phi} ,\quad \phi=(0,\tfrac\pi2) ,\\ S_{PAB}(\phi)&=b^2\,\tan\phi =\frac{r^2}{\sin^2\phi}\,\tan\phi =\frac{2\,r^2}{\sin2\,\phi} . \end{align}
Note that
\begin{align} \lim_{\phi\to0}S_{PAB}(\phi) &= \lim_{\phi\to\tfrac\pi2}S_{PAB}(\phi) =\infty \end{align}
\begin{align} S'_{PAB}(\phi) &=-\frac{4\,r^2\,\cos2\phi}{\sin^2(2\phi)} ,\\ S'_{PAB}(\phi):\quad\begin{cases} &<0,\quad \phi\in(0,\tfrac\pi4) ,\\ &=0,\quad \phi=\tfrac\pi4 ,\\ &>0 ,\quad \phi\in(\tfrac\pi4,\tfrac\pi2) \end{cases} . \end{align}
Thus the minimal area of $\triangle PAB$ is reached at $\phi=\tfrac\pi4$, hence $\angle BPA=\tfrac\pi2$, the center of its circumscribed circle is in the middle of $AB$ and circumradius $R=|OA|=|OB|=|OP|=r\sqrt2=\sqrt{32}$.
The only possibility that the center of the circumcenter is $(0,0)$ is that $b = OP = OA$, that is, equaling areas, $$\sqrt2\space b\cdot 4=b^2\iff b=4\sqrt2$$ so we have $$x^2+y^2=32$$ Thus it is the angle $\angle{APB}$ and not $\angle{PAB}$ which is equal to $90^{\circ}$.
Let $A(-a,0)$, $B(a, 0)$ and the radius of the given circle $r=4$. Then, the area of APB is
$$r\cdot PA = r \sqrt{a^2+b^2} \ge 2rab$$
where the equality, or, the minimum area, occurs at $a=b= \sqrt2 r=4\sqrt2$. Thus, APB is an isosceles right triangle whose circumcircle is centered at origin with the radius $a=b =4\sqrt2$, i.e.
$$x^2+y^2=32$$
