S straight line with slope 2 and y-intercept 5 touches the circle `x^2+y^2+16 x+12 y+c=0` at a point `Q` . Then the coordinates of `Q` are `(-6,11)` (b) `(-9,-13)` `(-10 ,-15)` (d) `(-6,-7)`

Options

A. `(-6,11)`
B. `(-9,-13)`
C. `(-10,-15)`
D. `(-6,-7)`

Explanation

To find the coordinates of the point \( Q \) where the straight line touches the circle, we will follow these steps:

### Step 1: Write the equation of the line
The equation of the line with slope \( m = 2 \) and y-intercept \( 5 \) can be written as:
\[ y = 2x + 5 \]

### Step 2: Write the equation of the circle
The given equation of the circle is:
\[ x^2 + y^2 + 16x + 12y + c = 0 \]
To find the center and radius, we can rewrite it in standard form. Completing the square for \( x \) and \( y \):
\[ x^2 + 16x + y^2 + 12y + c = 0 \]
Completing the square for \( x \):
\[ x^2 + 16x = (x + 8)^2 - 64 \]
Completing the square for \( y \):
\[ y^2 + 12y = (y + 6)^2 - 36 \]
Substituting back, we get:
\[ (x + 8)^2 - 64 + (y + 6)^2 - 36 + c = 0 \]
This simplifies to:
\[ (x + 8)^2 + (y + 6)^2 + (c - 100) = 0 \]
Thus, the center of the circle is \( (-8, -6) \).

### Step 3: Find the slope of the radius at point \( Q \)
Since the line is tangent to the circle at point \( Q \), the radius to point \( Q \) is perpendicular to the tangent line. The slope of the radius can be calculated as:
\[ \text{slope of radius} = \frac{y_1 + 6}{x_1 + 8} \]
where \( (x_1, y_1) \) are the coordinates of point \( Q \).

### Step 4: Set up the perpendicularity condition
Since the slope of the tangent line is \( 2 \), the product of the slopes of the tangent and radius must equal \(-1\):
\[ 2 \cdot \frac{y_1 + 6}{x_1 + 8} = -1 \]
This simplifies to:
\[ 2(y_1 + 6) = - (x_1 + 8) \]
or
\[ 2y_1 + 12 = -x_1 - 8 \]
Rearranging gives us:
\[ 2y_1 + x_1 + 20 = 0 \quad \text{(Equation 1)} \]

### Step 5: Substitute \( y_1 \) from the line equation
From the line equation \( y = 2x + 5 \), we can express \( y_1 \) in terms of \( x_1 \):
\[ y_1 = 2x_1 + 5 \quad \text{(Equation 2)} \]

### Step 6: Substitute \( y_1 \) into Equation 1
Substituting Equation 2 into Equation 1:
\[ 2(2x_1 + 5) + x_1 + 20 = 0 \]
This simplifies to:
\[ 4x_1 + 10 + x_1 + 20 = 0 \]
\[ 5x_1 + 30 = 0 \]
Solving for \( x_1 \):
\[ x_1 = -6 \]

### Step 7: Find \( y_1 \)
Substituting \( x_1 = -6 \) back into Equation 2 to find \( y_1 \):
\[ y_1 = 2(-6) + 5 = -12 + 5 = -7 \]

### Conclusion
Thus, the coordinates of point \( Q \) are:
\[ Q = (-6, -7) \]