The points $A,B,C$ are in this order on line $D$, and $AB = 4BC$. Let $M$ be a variable point on the perpendicular to $D$ through $C$. Let $MT_1$ and $MT_2$ be tangents to the circle with center $A$ and radius $AB$. Determine the locus of the orthocenter of the triangle $MT_1T_2.$
1. **Given Information and Setup:**
   - Points \( A, B, C \) are collinear on line \( D \) with \( AB = 4BC \).
   - \( M \) is a variable point on the perpendicular to \( D \) through \( C \).
   - \( MT_1 \) and \( MT_2 \) are tangents to the circle centered at \( A \) with radius \( AB \).

2. **Intersection and Orthocenter:**
   - Let \( AM \) intersect \( T_1T_2 \) at \( E \).
   - The orthocenter of \( \triangle AT_1T_2 \) is \( H \).

3. **Properties of the Rhombus:**
   - Since \( T_1H \perp AT_2 \) and \( AT_2 \perp AT_1 \), we have \( AT_2 \parallel T_1H \).
   - Thus, \( AT_2HT_1 \) forms a rhombus.
   - Since \( H \) lies on \( MA \), we have \( AH = 2AE \).

4. **Power of a Point:**
   - Using the power of a point theorem, \( AE \cdot AM = AT_2^2 = AB^2 \).
   - Therefore, \( AH \cdot AM = 2AB^2 \).

5. **Finding Point \( F \):**
   - Let \( F \) be on the extension of \( AC \) such that \( \frac{AF}{AB} = \frac{8}{5} \).
   - Then, \( AC \cdot AF = AB^2 \cdot \frac{5}{4} \cdot \frac{8}{5} = 2AB^2 = AH \cdot AM \).

6. **Cyclic Quadrilateral:**
   - Since \( AC \cdot AF = AH \cdot AM \), the points \( M, H, C, F \) are concyclic.
   - This implies \( \angle FHM = 90^\circ \).

7. **Conclusion:**
   - Since \( F \) is fixed, the locus of \( H \) is the circle with diameter \( AF \).

The final answer is the circle with diameter \( \boxed{ AF } \).