SOLUTION: The graph of the equation 4x^2 - 12x + 4y^2 + 16y - 15 = 10x + 25y + 28 is a circle. Find the radius of the circle.

Question 1209655: The graph of the equation
4x^2 - 12x + 4y^2 + 16y - 15 = 10x + 25y + 28
is a circle. Find the radius of the circle.

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
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 $\boxed{\frac{\sqrt{1253}}{8}}$

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