If (1,2) (2,a) are extremities of a diameter of the circle `x^(2)+y^(2)_3x-4y+6=0` then a=

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the value of \( a \) given that the points \( (1, 2) \) and \( (2, a) \) are the extremities of a diameter of the circle defined by the equation \( x^2 + y^2 - 3x - 4y + 6 = 0 \).

### Step 1: Identify the center of the circle
The general form of the circle equation is:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From the given equation \( x^2 + y^2 - 3x - 4y + 6 = 0 \), we can identify:
- \( 2g = -3 \) → \( g = -\frac{3}{2} \)
- \( 2f = -4 \) → \( f = -2 \)
- \( c = 6 \)

The center \( (h, k) \) of the circle can be found using:
\[ h = -g = \frac{3}{2}, \quad k = -f = 2 \]
Thus, the center of the circle is \( \left( \frac{3}{2}, 2 \right) \).

### Step 2: Find the midpoint of the diameter
The points \( (1, 2) \) and \( (2, a) \) are the endpoints of the diameter. The midpoint \( M \) of the diameter can be calculated as:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{1 + 2}{2}, \frac{2 + a}{2} \right) = \left( \frac{3}{2}, \frac{2 + a}{2} \right) \]

### Step 3: Set the midpoint equal to the center
Since the midpoint of the diameter is also the center of the circle, we can set the coordinates equal to the center:
\[ \frac{3}{2} = \frac{3}{2} \quad \text{(this is already satisfied)} \]
\[ 2 = \frac{2 + a}{2} \]

### Step 4: Solve for \( a \)
Now we solve the equation:
\[ 2 = \frac{2 + a}{2} \]
Multiplying both sides by 2 gives:
\[ 4 = 2 + a \]
Subtracting 2 from both sides results in:
\[ a = 4 \]

### Conclusion
Thus, the value of \( a \) is \( 4 \).