Question 1209655
Here's how to find the radius of the circle:

1. **Rewrite the equation by grouping x and y terms:**

4x² - 12x + 4y² + 16y - 15 = 10x + 25y + 28
4x² - 22x + 4y² - 9y = 43

2. **Complete the square for x and y:**

* **For x:**
  4(x² - (11/2)x)  
  To complete the square, take half of -11/2 which is -11/4 and square it to get 121/16.
  4(x² - (11/2)x + 121/16) - 4(121/16)

* **For y:**
  4(y² - (9/4)y)
  To complete the square, take half of -9/4 which is -9/8 and square it to get 81/64
  4(y² - (9/4)y + 81/64) - 4(81/64)

3. **Substitute the completed squares back into the equation:**

4(x² - (11/2)x + 121/16) - 4(121/16) + 4(y² - (9/4)y + 81/64) - 4(81/64) = 43
4(x - 11/4)² - 121/4 + 4(y - 9/8)² - 81/16 = 43

4. **Simplify and rewrite in standard circle form:**

Multiply the entire equation by 16 to eliminate fractions:

64(x - 11/4)² - 484 + 64(y - 9/8)² - 81 = 688
64(x - 11/4)² + 64(y - 9/8)² = 688 + 484 + 81
64(x - 11/4)² + 64(y - 9/8)² = 1253

Divide by 64:
(x - 11/4)² + (y - 9/8)² = 1253/64

5. **Identify the radius:**

The standard form of a circle's equation is (x - h)² + (y - k)² = r², where (h, k) is the center and r is the radius.

In our equation, r² = 1253/64. Therefore, the radius is:

r = √(1253/64) = √1253 / 8

Final Answer: The final answer is $\boxed{\frac{\sqrt{1253}}{8}}$