Yes, your attempt to the problem is correct. here is another way of solving it.
The equation of a circle is $$x^2 + y^2 = R^2\Rightarrow x^2 + y^2 = (\frac{d}2)^2$$
where, R and d are the radius and the diameter of the circle respectively.
So, the equation for the top semi circle is $$y = \sqrt{(\frac{d}2)^2 - x^2}$$
Let, the diameter of the circle be of the form $$d = 10a + b$$
where, $a$ and $b$ are whole numbers between 0 and 9 inclusive.
So, $a$ and $b$ are the digits of the two digit number.
Also, it is given that the length of cord CD is obtained by reversing the digits of the diameter. So, $$CD = 10b + a$$
Let us define a function $C(x)$ which gives us the length of a perpendicular chord, whose input is the distance of the center of the chord from the center of the circle.
We know that the length of half of the perpendicular chord is given by $$\sqrt{(\frac{d}2)^2 - x^2}$$
So,$$C(x) = 2\sqrt{(\frac{d}2)^2 - x^2}\Rightarrow 2\sqrt{(\frac{10a + b}2)^2 - x^2}$$
Now, let the distance of CD from the center of the circle be a rational number of the form $\frac{p}q$, where $p$ and $q$ are relatively prime integers.
Hence, the length of CD is given by $$C(\frac{p}q) = 2\sqrt{(\frac{10a + b}2)^2 - (\frac{p}q)^2}$$$$\Rightarrow10b + a = 2\sqrt{(\frac{10a + b}2)^2 - (\frac{p}q)^2}$$
Simplifying the equation will give us,$$(\frac{2p}{3q})^2 = 11(a^2-b^2)$$Since, the LHS is a perfect square, RHS must also be a perfect square. As, the RHS is already a multiple of $11$, $ a^2-b^2$ must be an odd power of 11, like $11^1, 11^2, 11^3$, etc.
Since, $a$ and $b$ are whole numbers between 0 and 9, the maximum value of $a^2-b^2$ will be when $a = 9$ and $b = 0$ , i.e. $81$.
Because, all odd powers of $11$ are greater than $81$ except $11$ itself, $a^2 - b^2$ must be $11$.
So, when we see a list of perfect squares from $1^2$ to $9^2$ $$1, 4, 9, 16, 25, 36, 49, 64, 81$$we see that only $36\;(i.e.\;6^2)$ and $25\;(i.e.\;5^2)$ are $11$ apart. Hence, we can choose, $a = 6$ and $b = 5$, to satisfy the equation $a^2 - b^2 = 11$.
Therefore, the length of the diameter of the circle is $65$ units.
You have $\overline{abc}=100a +10b+ c=37 (a+b+c)$ or
$$63a-27b-36c=0 \ \ \Longrightarrow 7a=3b+4c$$
Now $a \le \frac{3b+4c}{7}\le \frac{3\times 8+4 \times 9}{7}<9$.
Let $a=8$. Then $3b+4c=56$. We have $3b=4(14-c)$. Thus $b$ is a multiple of $4$. This forces that $b=4$ and gives $c=11$, which is impossible!
Let $a=7$. Then $3b+4c=49$ and $b=\frac{49-4c}{3}=16+\frac{1-4c}{3}$. This gives $c=4$ and leads to $b=11$, which is impossible again.
Let $a=6$. Then $3b+4c=42$ and $4c=3(14-b)$. This gives $c=6, 9$ since $c$ must be a multiple of $3$. Then possible pairs for $(b, c)$ are $(6, 6), (2, 9)$.
Thus largest is $629$.
Best Answer
I'd treat the original number as a single variable $x$. Then you have
\begin{align*} 10x+3 &= x+372 \\ 9x &= 369 \\ x &= 41 \end{align*}
As already pointed out in a comment, you apparently made some arithmetic mistake which led you to $43$ instead of $41$ in (your equivalent of) that final line.