what is $a^2+9=b^2+16=1+(a+b)^2$ solve for $a,b$ This is for a geometry question, and through a construction arrived at this equation. I could not solve it and after plugging it into wolfram got the correct answer but can anyone show a method for finding a,b
they are $\frac{5}{\sqrt3}$ and $\frac{2}{\sqrt3}$ respectively (note since this is a geometry question the lengths of a side cannot be negative so the negative solution sets do not matter)

Let
\begin{align} 
a^2+9&=k
\tag{1}\label{1}
,\\
b^2+16&=k
\tag{2}\label{2}
,\\
1+(a+b)^2&=k
\tag{3}\label{3}
\end{align}
for some $k>0$.
Then \eqref{3}$-$\eqref{1}$-$\eqref{2} gives
\begin{align}
ab &= 12-\tfrac12\,k
\tag{4}\label{4}
,\\
a^2b^2 &= (12-\tfrac12\,k)^2
\tag{5}\label{5}
.
\end{align}
On the other hand, from \eqref{1},\eqref{2}
we have
\begin{align}
a^2b^2 &= (k-9)(k-16)
\tag{6}\label{6}
.
\end{align}
Combination of \eqref{5} and \eqref{6}
results in equation in $k$:
\begin{align}
k(3k-52)&=0
\tag{7}\label{7}
,
\end{align}
and since $k>0$, we have the only option
\begin{align}
k&=\frac{52}3
\tag{8}\label{8}
.
\end{align}
Then \eqref{1} and \eqref{2} gives possible values of $a$
and $b$:
\begin{align}
a&=\pm \tfrac53\sqrt3
\tag{9}\label{9}
,\\
b&=\pm \tfrac23\sqrt3
\tag{10}\label{10}
.
\end{align}
And since $a,b$ also must be positive, the solution is ready.
