An arithmetic figure is comprising of two fonts. If we add both and square the digits and add 9 to it, the final figure is the same if we interchange the position of original figure. What is the final figure? i-e. Reflex the final figure.

Let's denote the two digits of the figure as \( a \) and \( b \). Then, the original figure can be represented as the two-digit number \( 10a + b \).

According to the problem, if we add both digits (\( a + b \)), square that sum, and then add 9, we should have the same result as interchanging the digits (which gives the number \( 10b + a \)).

We can express that mathematically as follows:

\[
(a + b)^2 + 9 = 10b + a
\]

Now, let's simplify this equation step by step.

1. Expand the left side:

\[
a^2 + 2ab + b^2 + 9 = 10b + a
\]

2. Rearranging gives us:

\[
a^2 + 2ab + b^2 - 10b - a + 9 = 0
\]

This is a quadratic in terms of \( a \):

\[
a^2 + (2b - 1)a + (b^2 - 10b + 9) = 0
\]

To find integer values for \( a \) and \( b \), we can apply the quadratic formula \( a = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \):

Here, \( A = 1 \), \( B = 2b - 1 \), and \( C = b^2 - 10b + 9 \).

The discriminant (inside the square root) should be a perfect square:

\[
\Delta = (2b - 1)^2 - 4(1)(b^2 - 10b + 9)
\]

Calculating \(\Delta\):

\[
\Delta = (4b^2 - 4b + 1) - (4b^2 - 40b + 36)
\]
\[
= 4b - 35
\]

For \(\Delta\) to be a perfect square, \( 4b - 35 \) must be a non-negative perfect square.

Let \( k^2 = 4b - 35 \).

Rearranging gives:

\[
4b = k^2 + 35 \implies b = \frac{k^2 + 35}{4}
\]

Next, since \( b \) must be a digit (0 to 9), we can try different integer values of \( k\).

1. \( k = 1 \): \( b = \frac{1 + 35}{4} = 9 \) → valid
2. \( k = 3 \): \( b = \frac{9 + 35}{4} = 11 \) → invalid
3. \( k = 5 \): \( b = \frac{25 + 35}{4} = 15 \) → invalid
4. \( k = 7 \): \( b = \frac{49 + 35}{4} = 21 \) → invalid
5. \( k = 0 \): \( k^2 = 0 \), \( b = \frac{35}{4} = 8.75 \) → invalid
6. \( k = 2, k = 4, k = 6 \): similar results; non-valid.

From \( k = 1 \), we found \( b = 9 \).

Substituting \( b = 9 \) back to see values of \( a \):

We need to verify with \( b = 9 \):
\[
(2(9) - 1)a + (9^2 - 10(9) + 9) = 0
\]

Calculating leads to:
\[
17a + 0 = 0 \rightarrow a = 0,
\]
which forms digit 09.

Thus, our original number is \( 09 \) or \( 9 \), and the final reflex figure is:

Final Figure Reflex: 9 → 9.