The polar of the line 8x-2y=11 with respect to the circle `2x^(2)+2y^(2)=11` is

Options

A. a) (4,1)
B. b) (4,-1)
C. c) (3,1)
D. d) (4,2)

Answer

To find the polar of the line \(8x - 2y = 11\) with respect to the circle \(2x^2 + 2y^2 = 11\), we can follow these steps:

### Step 1: Rewrite the Circle Equation
First, we simplify the equation of the circle:
\[ 2x^2 + 2y^2 = 11 \implies x^2 + y^2 = \frac{11}{2} \]
This is the standard form of the circle centered at the origin with radius \(\sqrt{\frac{11}{2}}\).

### Step 2: Identify the Line Equation
The line equation is given as:
\[ 8x - 2y = 11 \]
We can rewrite this in slope-intercept form:
\[ -2y = -8x + 11 \implies y = 4x - \frac{11}{2} \]

### Step 3: Find the Pole of the Line
To find the pole of the line with respect to the circle, we need to express the line in a suitable form. The general form of a line is \(Ax + By + C = 0\). Here, we can rewrite the line as:
\[ 8x - 2y - 11 = 0 \]
Thus, \(A = 8\), \(B = -2\), and \(C = -11\).

### Step 4: Use the Polar Equation Formula
The polar of a line \(Ax + By + C = 0\) with respect to the circle \(x^2 + y^2 = r^2\) is given by:
\[ \frac{Ax}{r^2} + \frac{By}{r^2} + \frac{C}{r^2} = 0 \]
In our case, \(r^2 = \frac{11}{2}\). Therefore, we have:
\[ \frac{8x}{\frac{11}{2}} + \frac{-2y}{\frac{11}{2}} + \frac{-11}{\frac{11}{2}} = 0 \]
Multiplying through by \(\frac{11}{2}\) to eliminate the denominators:
\[ 8x - 2y - 11 = 0 \]

### Step 5: Find the Chord of Contact
The chord of contact from the point \(P(h, k)\) to the circle is given by:
\[ xh + yk = r^2 \]
Substituting \(r^2 = \frac{11}{2}\):
\[ xh + yk = \frac{11}{2} \]

### Step 6: Set Up the System of Equations
Now we have two equations:
1. \(xh + yk = \frac{11}{2}\) (Chord of contact)
2. \(8x - 2y = 11\) (Line equation)

### Step 7: Solve for \(h\) and \(k\)
From the two equations, we can set up the ratios:
\[ \frac{h}{8} = \frac{k}{-2} = \frac{\frac{11}{2}}{11} \]
This implies:
\[ \frac{h}{8} = \frac{1}{2} \implies h = 4 \]
\[ \frac{k}{-2} = \frac{1}{2} \implies k = -1 \]

### Final Result
Thus, the pole of the line \(8x - 2y = 11\) with respect to the circle \(2x^2 + 2y^2 = 11\) is:
\[ (h, k) = (4, -1) \]

### Conclusion
The polar of the line with respect to the circle is given by the point \( (4, -1) \).