Let the tangents drawn to the circle x^2+y^2=16 from the point P(0,h) meet the x-axis at p…

Question

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 $	riangle APB$ is minimum, then $h$ is equal to:

Accepted Answer

With $OP=h$ and radius $a=4$, if $\theta$ is half the angle $\angle APB$ at $P$, one finds $OA=\dfrac{4}{\sin\theta}$ (using the geometry of the tangent triangle) and the area works out to
$$\text{Area}=\frac{32}{\sin2\theta},$$
which is minimized when $\sin2\theta=1$, i.e. $\theta=45^\circ$. Then $h=\dfrac{4}{\sin45^\circ}=4\sqrt2.$

Solutions · 1