Let $ABC$ be a triangle with $BC=30$, $AC=50$, and $AB=60$.  Circle $\omega_B$ is the circle passing through $A$ and $B$ tangent to $BC$ at $B$; $\omega_C$ is defined similarly.  Suppose the tangent to $\odot(ABC)$ at $A$ intersects $\omega_B$ and $\omega_C$ for the second time at $X$ and $Y$ respectively.  Compute $XY$.

Let $T = TNYWR$. For some positive integer $k$, a circle is drawn tangent to the coordinate axes such that the lines $x + y = k^2, x + y = (k+1)^2, \dots, x+y = (k+T)^2$ all pass through it. What is the minimum possible value of $k$?


### Part 1: Compute \( XY \)

1. **Identify the given elements and properties:**
   - Triangle \( ABC \) with sides \( BC = 30 \), \( AC = 50 \), and \( AB = 60 \).
   - Circle \( \omega_B \) passes through \( A \) and \( B \) and is tangent to \( BC \) at \( B \).
   - Circle \( \omega_C \) passes through \( A \) and \( C \) and is tangent to \( BC \) at \( C \).
   - The tangent to \( \odot(ABC) \) at \( A \) intersects \( \omega_B \) and \( \omega_C \) for the second time at \( X \) and \( Y \) respectively.

2. **Use the tangency properties:**
   - Since \( \omega_B \) is tangent to \( BC \) at \( B \), the tangency implies that \( \angle XAB \sim \angle BCA \).
   - Similarly, since \( \omega_C \) is tangent to \( BC \) at \( C \), the tangency implies that \( \angle YAC \sim \angle CBA \).

3. **Calculate the lengths \( XA \) and \( AY \):**
   - From the similarity \( \angle XAB \sim \angle BCA \), we can infer that \( XA \) is proportional to \( AB \) and \( BC \).
   - From the similarity \( \angle YAC \sim \angle CBA \), we can infer that \( AY \) is proportional to \( AC \) and \( BC \).

4. **Determine the lengths using the given side lengths:**
   - Given \( AB = 60 \), \( AC = 50 \), and \( BC = 30 \), we can use the proportionality to find \( XA \) and \( AY \).
   - Since \( \angle XAB \sim \angle BCA \), we have \( \frac{XA}{AB} = \frac{BC}{AC} \). Thus, \( XA = \frac{BC \cdot AB}{AC} = \frac{30 \cdot 60}{50} = 36 \).
   - Similarly, since \( \angle YAC \sim \angle CBA \), we have \( \frac{AY}{AC} = \frac{BC}{AB} \). Thus, \( AY = \frac{BC \cdot AC}{AB} = \frac{30 \cdot 50}{60} = 25 \).

5. **Sum the lengths to find \( XY \):**
   - \( XY = XA + AY = 36 + 25 = 61 \).

\[
\boxed{61}
\]

### Part 2: Find the minimum possible value of \( k \)

1. **Set up the problem:**
   - A circle is tangent to the coordinate axes and the lines \( x + y = k^2, x + y = (k+1)^2, \dots, x + y = (k+T)^2 \) all pass through it.
   - Let the center of the circle be \( (a, a) \).

2. **Use the distance from the center to the lines:**
   - The distance from the center \( (a, a) \) to the line \( x + y = k^2 \) is \( \frac{|2a - k^2|}{\sqrt{2}} \).
   - Since the circle is tangent to the coordinate axes, its radius is \( a \).

3. **Set up the inequalities:**
   - For the farthest line \( x + y = (k+T)^2 \), the distance must be less than or equal to the radius:
     \[
     \frac{|2a - (k+T)^2|}{\sqrt{2}} \le a
     \]
   - For the nearest line \( x + y = k^2 \), the distance must be less than or equal to the radius:
     \[
     \frac{|2a - k^2|}{\sqrt{2}} \le a
     \]

4. **Combine the inequalities:**
   - Simplify the inequalities to find the relationship between \( k \) and \( T \):
     \[
     (2 - \sqrt{2})a \le k^2 \quad \text{and} \quad (k+T)^2 \le (2 + \sqrt{2})a
     \]
   - Combining these, we get:
     \[
     (2 - \sqrt{2})(k+T)^2 \le (2 + \sqrt{2})k^2
     \]

5. **Solve for \( k \):**
   - Simplify the inequality:
     \[
     \frac{(k+T)^2}{k^2} \le \frac{2 + \sqrt{2}}{2 - \sqrt{2}}
     \]
   - Using the approximation \( \frac{2 + \sqrt{2}}{2 - \sqrt{2}} \approx 3 + 2\sqrt{2} \), we get:
     \[
     \frac{(k+T)^2}{k^2} \le 3 + 2\sqrt{2}
     \]
   - Solving for \( k \), we find:
     \[
     k \ge \frac{T}{\sqrt{3 + 2\sqrt{2}} - 1}
     \]
   - For the minimum \( k \), we approximate and find \( k \approx 44 \).

\[
\boxed{44}
\]