1. Given line

   $x + by + c = 0$

   We need the triangle formed by this line with the coordinate axes.

2. Intercepts of the line

   • On the $x$-axis, $y=0$: $x + c = 0 \Rightarrow x = -c$
   • On the $y$-axis, $x=0$: $by + c = 0 \Rightarrow y = -\frac{c}{b}$

   So the area of the triangle formed with the axes is $\frac{1}{2}\left|(-c)\left(-\frac{c}{b}\right)\right| = 48$

   Hence, $\frac{1}{2}\left|\frac{c^2}{b}\right| = 48$ $\left|\frac{c^2}{b}\right| = 96$ $c^2 = 96|b|$

3. Condition on the perpendicular from origin

   For the line $x + by + c = 0,$ a normal vector is $(1,b)$.

   The perpendicular from the origin to the line is along this normal, so its slope is $b$.

   It makes an angle $45^\circ$ with the positive $x$-axis, so $\tan 45^\circ = 1 \Rightarrow b = 1$

   (If the line had made angle $135^\circ$, then slope would be $-1$, but the question says $45^\circ$ with positive $x$-axis.)

4. Find $c^2$

   Using $b=1$ in $c^2 = 96|b|,$ we get $c^2 = 96$

5. Compute $b^2 + c^2$

   $b^2 + c^2 = 1^2 + 96 = 97$

6. Check options

   • A: $90$ ❌
   • B: $83$ ❌
   • C: $93$ ❌
   • D: $97$ ✅

Therefore, the correct answer is D.