P and Q are two symmetrical points about the tangent at origin to the circle `x^(2)+y^(2)-x+y=0`. If P be (-5,6), then Q is

Options

A. (6, 5)
B. (5, 6)
C. (6, -5)
D. (-6, 5)

Answer

To solve the problem, we need to find the coordinates of point Q, which is symmetric to point P about the tangent line to the circle at the origin. Let's go through the steps systematically.

### Step 1: Rewrite the equation of the circle
The given equation of the circle is:
\[ x^2 + y^2 - x + y = 0 \]

We can rearrange this equation:
\[ x^2 - x + y^2 + y = 0 \]
To complete the square, we rewrite it as:
\[ (x - \frac{1}{2})^2 + (y + \frac{1}{2})^2 = \frac{1}{2} \]

### Step 2: Identify the center and radius of the circle
From the completed square form, we can identify:
- Center \( C \left( \frac{1}{2}, -\frac{1}{2} \right) \)
- Radius \( r = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \)

### Step 3: Find the equation of the tangent line at the origin
The tangent line at the origin can be expressed as:
\[ y = mx \]
where \( m \) is the slope of the tangent.

### Step 4: Calculate the perpendicular distance from the center to the tangent line
The formula for the perpendicular distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

In our case, the tangent line can be rewritten as:
\[ mx - y = 0 \]
So, \( A = m, B = -1, C = 0 \).

The center of the circle is \( C \left( \frac{1}{2}, -\frac{1}{2} \right) \). Plugging these values into the distance formula:
\[ d = \frac{|m \cdot \frac{1}{2} - (-\frac{1}{2})|}{\sqrt{m^2 + 1}} = \frac{|m \cdot \frac{1}{2} + \frac{1}{2}|}{\sqrt{m^2 + 1}} \]

### Step 5: Set the distance equal to the radius
Since the distance from the center to the tangent line must equal the radius:
\[ \frac{|m \cdot \frac{1}{2} + \frac{1}{2}|}{\sqrt{m^2 + 1}} = \frac{1}{\sqrt{2}} \]

### Step 6: Solve for \( m \)
Squaring both sides and simplifying will yield a quadratic equation in \( m \). After solving, we find that \( m = 1 \) (the slope of the tangent line).

### Step 7: Write the equation of the tangent line
Thus, the equation of the tangent line is:
\[ y = x \]

### Step 8: Find the coordinates of point Q
Given point P is \( (-5, 6) \), we need to find point Q, which is symmetric to P about the line \( y = x \). To find the symmetric point, we can switch the coordinates of P:
- If P is \( (-5, 6) \), then Q will be \( (6, -5) \).

### Final Answer
Thus, the coordinates of point Q are:
\[ Q = (6, -5) \]