# Equation of circumcircle of a triangle which has minimum area

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 ?

• It is impossible that $\angle{PAB}=90^{\circ}$ Mar 15 '20 at 14:38

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.

$$x^2+y^2=32$$