Let `(alpha,beta)` be the circumcenter of the triangle formed by the lines
`4x+3y=69`,
`4y-3x=17`,and
`x+7y=61`
Then `(alpha-beta)^(2)+alpha+beta ` is equal to

Options

A. `16`
B. `15`
C. `18`
D. `17`

Answer

To solve the problem, we need to find the circumcenter of the triangle formed by the given lines and then calculate the expression \((\alpha - \beta)^2 + \alpha + \beta\).

### Step 1: Find the intersection points of the lines

We have three lines:
1. \(4x + 3y = 69\) (Line 1)
2. \(4y - 3x = 17\) (Line 2)
3. \(x + 7y = 61\) (Line 3)

**Finding the intersection of Line 1 and Line 2:**

From Line 1:
\[ 3y = 69 - 4x \]
\[ y = \frac{69 - 4x}{3} \]

Substituting \(y\) in Line 2:
\[ 4\left(\frac{69 - 4x}{3}\right) - 3x = 17 \]
Multiplying through by 3 to eliminate the fraction:
\[ 4(69 - 4x) - 9x = 51 \]
\[ 276 - 16x - 9x = 51 \]
\[ 276 - 25x = 51 \]
\[ 25x = 276 - 51 \]
\[ 25x = 225 \]
\[ x = 9 \]

Now substituting \(x = 9\) back into Line 1 to find \(y\):
\[ 4(9) + 3y = 69 \]
\[ 36 + 3y = 69 \]
\[ 3y = 33 \]
\[ y = 11 \]

Thus, the intersection point \(A\) is \((9, 11)\).

**Finding the intersection of Line 1 and Line 3:**

Substituting \(y\) from Line 3 into Line 1:
\[ 4x + 3\left(\frac{61 - x}{7}\right) = 69 \]
Multiplying through by 7 to eliminate the fraction:
\[ 28x + 3(61 - x) = 483 \]
\[ 28x + 183 - 3x = 483 \]
\[ 25x = 300 \]
\[ x = 12 \]

Now substituting \(x = 12\) back into Line 3 to find \(y\):
\[ 12 + 7y = 61 \]
\[ 7y = 49 \]
\[ y = 7 \]

Thus, the intersection point \(B\) is \((12, 7)\).

**Finding the intersection of Line 2 and Line 3:**

Substituting \(y\) from Line 3 into Line 2:
\[ 4y - 3x = 17 \]
Substituting \(y = \frac{61 - x}{7}\):
\[ 4\left(\frac{61 - x}{7}\right) - 3x = 17 \]
Multiplying through by 7:
\[ 4(61 - x) - 21x = 119 \]
\[ 244 - 4x - 21x = 119 \]
\[ 244 - 25x = 119 \]
\[ 25x = 125 \]
\[ x = 5 \]

Now substituting \(x = 5\) back into Line 3 to find \(y\):
\[ 5 + 7y = 61 \]
\[ 7y = 56 \]
\[ y = 8 \]

Thus, the intersection point \(C\) is \((5, 8)\).

### Step 2: Find the circumcenter of triangle ABC

The circumcenter of a right triangle lies at the midpoint of the hypotenuse. Here, we can find the midpoint of line segment \(AB\) (since \(AB\) is the hypotenuse).

Coordinates of points:
- \(A(9, 11)\)
- \(B(12, 7)\)

Midpoint \(M\) of \(AB\):
\[ M = \left(\frac{9 + 12}{2}, \frac{11 + 7}{2}\right) = \left(\frac{21}{2}, \frac{18}{2}\right) = \left(\frac{21}{2}, 9\right) \]

Thus, \(\alpha = \frac{21}{2}\) and \(\beta = 9\).

### Step 3: Calculate \((\alpha - \beta)^2 + \alpha + \beta\)

Calculating \(\alpha - \beta\):
\[ \alpha - \beta = \frac{21}{2} - 9 = \frac{21}{2} - \frac{18}{2} = \frac{3}{2} \]

Calculating \((\alpha - \beta)^2\):
\[ (\alpha - \beta)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \]

Calculating \(\alpha + \beta\):
\[ \alpha + \beta = \frac{21}{2} + 9 = \frac{21}{2} + \frac{18}{2} = \frac{39}{2} \]

Now, combining these results:
\[ (\alpha - \beta)^2 + \alpha + \beta = \frac{9}{4} + \frac{39}{2} \]
Converting \(\frac{39}{2}\) to quarters:
\[ \frac{39}{2} = \frac{78}{4} \]
Thus,
\[ \frac{9}{4} + \frac{78}{4} = \frac{87}{4} \]

### Final Result
The value of \((\alpha - \beta)^2 + \alpha + \beta\) is \(\frac{87}{4}\).