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.
"Here I am stuck as to what I can do. One option is just checking all the squares from 32→99 according to the needed conditions. " Well, don't check the odd ones. And don't check $\sqrt{1000} \le n^2 \le \sqrt {1999}$ or $\sqrt{8889}\le n^2 \le \sqrt{9999}$ so that tells us to only check $46$ through $94$.
Let $n = 10a + b$ then $(10a + b)= 100a^2 + 20ab + b^2$
$b^2 = 0,4,16,36,64$. Now if $b = 4$ or $6$ then $b^2$ will cause an odd digit to be carried of the the tens place. And the tens place will be determined by $2ab$ plus the odd number added. This results in an odd number. So that is impossible.
$b^2 = 0,4,$ or $64$ and $b= 0,2,$ or $b=8$.
If $b = 0$ then we need $a^2=0,4,16,36,64$ to be a perfect square with two even digits. That can only be $a = 8$. So
So $80^2 = 6400$ is one such number.
Now we just need to check $48, 52,58, 62,68, 72,78,82,88,92$.
But $2889...3999$ all have od digits so we don't have to check $53.. 63$. Or $\sqrt {4889}...\sqrt{5999}$ or $70... 77$. Or $\sqrt{6889}...{7999}$ or $83..89$.
So we only need to check $48,52,78,82,92$
$(10a + 2)^2 = 100a^2 + 40a + 4$ which means the value carried by $4a$ whether odd or even must make $a^2$ odd of even. so the digit carried by $4a$ and an d $a$ must be the same parity.
$4\cdot 5=20$ so $2$ is even but $5$ is odd. So $(50+2)^2 = 2500 + 40\cdot 5 + 4$ so the $2$ carried by $40\cdot 5$ to $2500$ will make $2700$ i.e $2704$.
$4\cdot 8 =32$. The $3$ is odd but $8$ is even so the first two digits of $82^2$ will be $8^2 +3$ which is odd.
$4\cdot 9=36$ and $3$ is odd as is $9$ so this is good. $92^2 = 8100 + 360 + 4=8464$ with the $3$ and 81$ combining to make an even.
That just leaves $48$ and $78$ to check. It's easier to just check them then to make a carrying rule. $48^2 = 2304$. Nope. and $78^2=6084$. Good.
So $80^2 = 6400,92^2 = 8464$ and $78^2 = 6084$ are the only $3$.
Best Answer
Hint $\,{\rm mod}\ 37\!:\ \color{#c00}{10^3\equiv 1}\,$ so multiplication by $10\,$ does a cyclic shift $\, abc \mapsto bca,\,$ i.e.
$$\begin{eqnarray} abc_{\:\!10} &=&\, 10^2 a +\, 10\:\!\ b + c\\[.3em] \Longrightarrow\ 10\,abc_{\:\!10} &\equiv& \underbrace{\color{#c00}{10^3}}_{\large\color{#c00}1} a + 10^2 b + 10 c \,\equiv\, bca_{\:\!10}\end{eqnarray}\qquad$$
Therefore the problem reduces to counting the number of multiples of $\,37\,$ in the interval $\,[37000,37999].\,$ But your count is one low (a fencepost error).