Let $O$ and $A$ be two points in the plane with $OA = 30$, and let $\Gamma$ be a circle with center $O$ and radius $r$. Suppose that there exist two points $B$ and $C$ on $\Gamma$ with $\angle ABC = 90^{\circ}$ and $AB = BC$. Compute the minimum possible value of $\lfloor r \rfloor.$
1. **Fixing the triangle $\triangle ABC$**: We start by fixing the triangle $\triangle ABC$ such that $AB = BC = 2$. This is a valid assumption because we can scale the figure later to match the given conditions.

2. **Locating point $O$**: Since $O$ is the center of the circle $\Gamma$ and $B$ and $C$ lie on $\Gamma$, $O$ must lie on the perpendicular bisector of $BC$. Let the midpoint of $BC$ be $Y$. The length $OY$ is denoted as $y$.

3. **Calculating $OC$ and $OA$**: 
   - Since $Y$ is the midpoint of $BC$, $BY = YC = 1$.
   - Using the Pythagorean theorem in $\triangle OYC$, we get:
     \[
     OC = \sqrt{OY^2 + YC^2} = \sqrt{y^2 + 1}
     \]
   - Similarly, using the Pythagorean theorem in $\triangle OYA$, we get:
     \[
     OA = \sqrt{OY^2 + YA^2} = \sqrt{y^2 + 4y + 5}
     \]

4. **Minimizing $\frac{OC}{OA}$**: We need to minimize the ratio $\frac{OC}{OA}$, which is equivalent to minimizing:
   \[
   \frac{\sqrt{y^2 + 1}}{\sqrt{y^2 + 4y + 5}}
   \]
   This is the same as maximizing:
   \[
   \frac{y^2 + 4y + 5}{y^2 + 1}
   \]

5. **Maximizing the ratio**: To maximize $\frac{y^2 + 4y + 5}{y^2 + 1}$, we can use calculus. Let:
   \[
   f(y) = \frac{y^2 + 4y + 5}{y^2 + 1}
   \]
   We find the derivative $f'(y)$ and set it to zero to find the critical points:
   \[
   f'(y) = \frac{(2y + 4)(y^2 + 1) - (y^2 + 4y + 5)(2y)}{(y^2 + 1)^2}
   \]
   Simplifying the numerator:
   \[
   (2y + 4)(y^2 + 1) - (y^2 + 4y + 5)(2y) = 2y^3 + 2y + 4y^2 + 4 - 2y^3 - 8y^2 - 10y = -4y^2 - 8y + 4
   \]
   Setting the numerator to zero:
   \[
   -4y^2 - 8y + 4 = 0 \implies y^2 + 2y - 1 = 0 \implies y = -1 \pm \sqrt{2}
   \]
   Since $y$ must be positive, we take $y = \sqrt{2} - 1$.

6. **Calculating $OC$ and $OA$ with $y = \sqrt{2} - 1$**:
   \[
   OC = \sqrt{(\sqrt{2} - 1)^2 + 1} = \sqrt{2 - 2\sqrt{2} + 1 + 1} = \sqrt{4 - 2\sqrt{2}}
   \]
   \[
   OA = \sqrt{(\sqrt{2} - 1)^2 + 4(\sqrt{2} - 1) + 5} = \sqrt{2 - 2\sqrt{2} + 1 + 4\sqrt{2} - 4 + 5} = \sqrt{4 + 2\sqrt{2}}
   \]

7. **Finding the ratio**:
   \[
   \frac{OC}{OA} = \sqrt{\frac{4 - 2\sqrt{2}}{4 + 2\sqrt{2}}} = \sqrt{\frac{(4 - 2\sqrt{2})^2}{(4 + 2\sqrt{2})(4 - 2\sqrt{2})}} = \frac{4 - 2\sqrt{2}}{\sqrt{8}} = \sqrt{2} - 1
   \]

8. **Scaling to match $OA = 30$**:
   \[
   OA = 30 \implies OC = 30(\sqrt{2} - 1)
   \]
   Therefore, $r = 30(\sqrt{2} - 1)$.

9. **Finding the minimum possible value of $\lfloor r \rfloor$**:
   \[
   \lfloor 30(\sqrt{2} - 1) \rfloor = \lfloor 30 \times 0.414 \rfloor = \lfloor 12.42 \rfloor = 12
   \]

The final answer is $\boxed{12}$.