The point on the straight line `y = 2x + 11` which is nearest to the circle `16 (x^2 + y^2) + 32 x - 8y - 50 = 0` is

Options

A. (9/2, 2)
B. (-9/2, 2)
C. (9/2, -2)
D. none of these

Answer

To find the point on the straight line \( y = 2x + 11 \) that is nearest to the circle given by the equation \( 16(x^2 + y^2) + 32x - 8y - 50 = 0 \), we can follow these steps:

### Step 1: Rewrite the Circle Equation
First, we simplify the equation of the circle. The given equation is:

\[ 16(x^2 + y^2) + 32x - 8y - 50 = 0 \]

Dividing the entire equation by 16:

\[ x^2 + y^2 + 2x - \frac{1}{2}y - \frac{25}{8} = 0 \]

### Step 2: Rearranging to Standard Form
Rearranging gives:

\[ x^2 + 2x + y^2 - \frac{1}{2}y = \frac{25}{8} \]

Now, we complete the square for \( x \) and \( y \):

- For \( x^2 + 2x \):
\[ (x + 1)^2 - 1 \]

- For \( y^2 - \frac{1}{2}y \):
\[ \left(y - \frac{1}{4}\right)^2 - \frac{1}{16} \]

Substituting these back into the equation gives:

\[ (x + 1)^2 - 1 + \left(y - \frac{1}{4}\right)^2 - \frac{1}{16} = \frac{25}{8} \]

### Step 3: Combine and Simplify
Combining the constants:

\[ (x + 1)^2 + \left(y - \frac{1}{4}\right)^2 = \frac{25}{8} + 1 + \frac{1}{16} \]

Calculating the right side:

\[ \frac{25}{8} + \frac{8}{8} + \frac{1}{16} = \frac{33}{8} + \frac{1}{16} = \frac{66 + 1}{16} = \frac{67}{16} \]

Thus, the equation of the circle is:

\[ (x + 1)^2 + \left(y - \frac{1}{4}\right)^2 = \frac{67}{16} \]

### Step 4: Identify the Center and Radius
From the equation, the center \( C \) of the circle is \( (-1, \frac{1}{4}) \) and the radius \( r \) is \( \sqrt{\frac{67}{16}} = \frac{\sqrt{67}}{4} \).

### Step 5: Parameterize the Line
The line is given by \( y = 2x + 11 \). We can express a point \( A \) on this line as:

\[ A(k) = (k, 2k + 11) \]

### Step 6: Find the Slope of Line AC
The slope of line segment \( AC \) is given by:

\[ \text{slope of } AC = \frac{(2k + 11) - \frac{1}{4}}{k + 1} = \frac{2k + \frac{44}{4} - \frac{1}{4}}{k + 1} = \frac{2k + \frac{43}{4}}{k + 1} \]

### Step 7: Set Up the Perpendicular Condition
Since the slope of the tangent at point \( C \) is \( 2 \) (the slope of the line), the product of the slopes must equal \(-1\):

\[ \frac{2k + \frac{43}{4}}{k + 1} \cdot 2 = -1 \]

### Step 8: Solve for \( k \)
This gives us:

\[ (2k + \frac{43}{4}) \cdot 2 = - (k + 1) \]

Expanding and rearranging:

\[ 4k + \frac{43}{2} = -k - 1 \]

Combining like terms:

\[ 5k = -1 - \frac{43}{2} = -\frac{2 + 43}{2} = -\frac{45}{2} \]

Thus,

\[ k = -\frac{9}{2} \]

### Step 9: Find the Coordinates of Point A
Substituting \( k \) back into the parameterization of the line:

\[ A\left(-\frac{9}{2}\right) = \left(-\frac{9}{2}, 2\left(-\frac{9}{2}\right) + 11\right) = \left(-\frac{9}{2}, -9 + 11\right) = \left(-\frac{9}{2}, 2\right) \]

### Final Answer
The point on the straight line \( y = 2x + 11 \) that is nearest to the circle is:

\[ \boxed{\left(-\frac{9}{2}, 2\right)} \]