Point on the circle `x^(2)+y^(2)-2x+4y-4=0` which is nearest to the line `y=2x+11` is :

Options

A. `(1-(6)/(sqrt(5)), -2+(3)/(sqrt(5)))`
B. `(1+(6)/(sqrt(5)), 2-(3)/(sqrt(5)))`
C. `(1-(6)/(sqrt(5)), -2-(3)/(sqrt(5)))`
D. None of these

Answer

To find the point on the circle \( x^2 + y^2 - 2x + 4y - 4 = 0 \) that is nearest to the line \( y = 2x + 11 \), we can follow these steps:

### Step 1: Rewrite the Circle's Equation
We start by rewriting the equation of the circle in standard form.

The given equation is:
\[ x^2 + y^2 - 2x + 4y - 4 = 0 \]

We can rearrange it as:
\[ x^2 - 2x + y^2 + 4y = 4 \]

Next, we complete the square for \( x \) and \( y \).

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

For \( y^2 + 4y \):
\[ y^2 + 4y = (y + 2)^2 - 4 \]

Substituting back, we have:
\[ (x - 1)^2 - 1 + (y + 2)^2 - 4 = 4 \]

Simplifying gives:
\[ (x - 1)^2 + (y + 2)^2 = 9 \]

This shows that the center of the circle is \( (1, -2) \) and the radius is \( 3 \).

### Step 2: Find the Perpendicular Distance from the Center to the Line
The line is given by:
\[ y = 2x + 11 \]
This can be rewritten in standard form as:
\[ 2x - y + 11 = 0 \]

The center of the circle is \( (1, -2) \). We can use the formula for the distance \( d \) from a point \( (x_0, y_0) \) to a line \( Ax + By + C = 0 \):
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

Here, \( A = 2 \), \( B = -1 \), \( C = 11 \), and \( (x_0, y_0) = (1, -2) \):
\[ d = \frac{|2(1) - (-2) + 11|}{\sqrt{2^2 + (-1)^2}} = \frac{|2 + 2 + 11|}{\sqrt{4 + 1}} = \frac{15}{\sqrt{5}} = 3\sqrt{5} \]

### Step 3: Determine the Nearest Point on the Line
Since the radius of the circle is \( 3 \), we need to find the point on the line that is \( 3 \) units away from the center \( (1, -2) \) in the direction perpendicular to the line.

The slope of the line is \( 2 \), so the slope of the perpendicular line is \( -\frac{1}{2} \).

Using the point-slope form of the line:
\[ y + 2 = -\frac{1}{2}(x - 1) \]

This simplifies to:
\[ y = -\frac{1}{2}x - 1.5 \]

### Step 4: Find the Intersection Points
Now, we need to find the intersection points of this line with the original line \( y = 2x + 11 \). Setting the equations equal:
\[ -\frac{1}{2}x - 1.5 = 2x + 11 \]

Multiplying through by \( -2 \) to eliminate the fraction:
\[ x + 3 = -4x - 22 \]
\[ 5x = -25 \implies x = -5 \]

Substituting \( x = -5 \) back into the line equation:
\[ y = 2(-5) + 11 = -10 + 11 = 1 \]

Thus, one intersection point is \( (-5, 1) \).

### Step 5: Find the Point on the Circle
Now we need to find the point on the circle that is \( 3 \) units away from the center in the direction of the line. The direction vector from the center \( (1, -2) \) to the point \( (-5, 1) \) is:
\[ (-5 - 1, 1 + 2) = (-6, 3) \]

The unit vector in this direction is:
\[ \left(-\frac{6}{\sqrt{45}}, \frac{3}{\sqrt{45}}\right) = \left(-\frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right) \]

Now, we find the point on the circle:
\[ (1, -2) + 3\left(-\frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right) = \left(1 - \frac{6}{\sqrt{5}}, -2 + \frac{3}{\sqrt{5}}\right) \]

### Final Answer
The point on the circle nearest to the line \( y = 2x + 11 \) is:
\[ \left(1 - \frac{6}{\sqrt{5}}, -2 + \frac{3}{\sqrt{5}}\right) \]