Let `y=x+2,4y=3x+6` and `3y=4x+1` be three tangent lines to the circle `(x-h)^(2)+(y-k)^(2)=r^(2)`. Then `h+k` is equal to :

Options

A. `5`
B. `5(1+sqrt(2))`
C. `6`
D. `5sqrt(2)`

Answer

To solve the problem, we need to find the value of \( h + k \) where the lines given are tangents to the circle defined by the equation \( (x-h)^2 + (y-k)^2 = r^2 \).

### Step-by-step Solution:

1. **Identify the Tangent Lines**:
The equations of the tangent lines are given as:
- Line 1: \( y = x + 2 \) (which can be rewritten as \( x - y + 2 = 0 \))
- Line 2: \( 4y = 3x + 6 \) (which can be rewritten as \( 3x - 4y + 6 = 0 \))
- Line 3: \( 3y = 4x + 1 \) (which can be rewritten as \( 4x - 3y + 1 = 0 \))

2. **Find the Intersection of the Lines**:
We need to find the point where these lines intersect. We can solve the equations of two lines at a time.

Let's solve Line 1 and Line 2:
- From Line 1: \( y = x + 2 \)
- Substitute into Line 2:
\[ 3x - 4(x + 2) + 6 = 0 \\ 3x - 4x - 8 + 6 = 0 \\ -x - 2 = 0 \\ x = -2 \]
- Substitute \( x = -2 \) back into Line 1:
\[ y = -2 + 2 = 0 \]
- So, the intersection point of Line 1 and Line 2 is \( (-2, 0) \).

3. **Find the Intersection of Line 2 and Line 3**:
Now, we solve Line 2 and Line 3:
- From Line 3: \( 3y = 4x + 1 \) or \( y = \frac{4}{3}x + \frac{1}{3} \)
- Substitute into Line 2:
\[ 3x - 4\left(\frac{4}{3}x + \frac{1}{3}\right) + 6 = 0 \\ 3x - \frac{16}{3}x - \frac{4}{3} + 6 = 0 \\ \left(3 - \frac{16}{3}\right)x + 6 - \frac{4}{3} = 0 \\ \left(\frac{9}{3} - \frac{16}{3}\right)x + \frac{18}{3} - \frac{4}{3} = 0 \\ -\frac{7}{3}x + \frac{14}{3} = 0 \\ x = 2 \]
- Substitute \( x = 2 \) back into Line 3:
\[ y = \frac{4}{3}(2) + \frac{1}{3} = \frac{8}{3} + \frac{1}{3} = 3 \]
- So, the intersection point of Line 2 and Line 3 is \( (2, 3) \).

4. **Find the Intersection of Line 1 and Line 3**:
Now, we solve Line 1 and Line 3:
- Substitute \( y = x + 2 \) into Line 3:
\[ 4x - 3(x + 2) + 1 = 0 \\ 4x - 3x - 6 + 1 = 0 \\ x - 5 = 0 \\ x = 5 \]
- Substitute \( x = 5 \) back into Line 1:
\[ y = 5 + 2 = 7 \]
- So, the intersection point of Line 1 and Line 3 is \( (5, 7) \).

5. **Finding the Center of the Circle**:
The center of the circle lies on the angle bisectors of the tangents. The angle bisector of two lines can be found using the formula for the angle bisector. Here, we can find the midpoint of the intersection points \( (-2, 0) \), \( (2, 3) \), and \( (5, 7) \).

The average of the x-coordinates:
\[ h = \frac{-2 + 2 + 5}{3} = \frac{5}{3} \]
The average of the y-coordinates:
\[ k = \frac{0 + 3 + 7}{3} = \frac{10}{3} \]

6. **Calculate \( h + k \)**:
\[ h + k = \frac{5}{3} + \frac{10}{3} = \frac{15}{3} = 5 \]

### Final Answer:
Thus, \( h + k = 5 \).