Answer
$\frac{-\pi h+\sqrt{{{\pi }^{2}}{{h}^{2}}+2\pi A}}{2\pi }$.
Work Step by Step
$A=2\pi {{r}^{2}}+2\pi rh$
Subtract $A$ from both sides of the equation.
$\begin{align}
& A=2\pi {{r}^{2}}+2\pi rh \\
& A-A=2\pi {{r}^{2}}+2\pi rh-A \\
& 0=2\pi {{r}^{2}}+2\pi rh-A
\end{align}$
$2\pi {{r}^{2}}+2\pi rh-A=0$
Now, using the quadratic formula,
$r=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substitute the values in the equation,
$a=2\pi ,b=2\pi h,c=-A$
Solve the equation,
$\begin{align}
& r=\frac{-2\pi h\pm \sqrt{{{\left( 2\pi h \right)}^{2}}-4\times 2\pi \times \left( -A \right)}}{2\times 2\pi } \\
& =\frac{-2\pi h\pm \sqrt{4{{\pi }^{2}}{{h}^{2}}+8\pi A}}{4\pi } \\
& =\frac{-\pi h\pm \sqrt{{{\pi }^{2}}{{h}^{2}}+2\pi A}}{2\pi }
\end{align}$
So,
$r=\frac{-\pi h+\sqrt{{{\pi }^{2}}{{h}^{2}}+2\pi A}}{2\pi }$
The diameter of a circle is always twice its radius. Therefore, substitute r with $\frac{d}{2}$.
$\frac{d}{2}=\frac{-\pi h+\sqrt{{{\pi }^{2}}{{h}^{2}}+2\pi A}}{2\pi }$
$\begin{align}
& 2\times \frac{d}{2}=2\times \left( \frac{-\pi h+\sqrt{{{\pi }^{2}}{{h}^{2}}+2\pi A}}{2\pi } \right) \\
& d=\frac{-\pi h+\sqrt{{{\pi }^{2}}{{h}^{2}}+2\pi A}}{\pi }
\end{align}$
Thus, the value is $d=\frac{-\pi h+\sqrt{{{\pi }^{2}}{{h}^{2}}+2\pi A}}{\pi }$.
