Question

If the pole of the line \(x + 2by - 5 = 0\) with respect to the circle \(S = x^2 + y^2 - 4x - 6y + 4 = 0\) lies on the line \(x + by + 1 = 0\), then the polar of the point \((b, -b)\) with respect to the circle \(S = 0\) is

• A. \(5y - 6 = 0\)
• B. \(y - 6 = 0\)
• C. \(x + 5y - 6 = 0\)
• D. \(5x + y - 6 = 0\)

Hint:

To find the pole of a line \(Lx+My+N=0\) with respect to a circle \(x^2+y^2+2gx+2fy+c=0\), equate the coefficients of the given line with the general equation of the polar \(xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0\). This leads to a system of equations that can be solved for the pole \((x_1, y_1)\). Remember that if the pole lies on another line, its coordinates must satisfy the equation of that line.

Answer 1

The Correct Option is A

Step 1: Find the pole of the given line with respect to the circle \(S = 0\).
The equation of the circle is \(S = x^2 + y^2 - 4x - 6y + 4 = 0\).
Comparing with \(x^2 + y^2 + 2gx + 2fy + c = 0\), we have:
\(2g = -4 \implies g = -2\)
\(2f = -6 \implies f = -3\)
\(c = 4\)
Let the pole of the line \(x + 2by - 5 = 0\) with respect to the circle \(S = 0\) be \((x_1, y_1)\).
The polar of \((x_1, y_1)\) is \(xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0\).
Substituting the values of \(g, f, c\):
\(xx_1 + yy_1 - 2(x+x_1) - 3(y+y_1) + 4 = 0\)
Rearrange terms to match the form of a linear equation:
\(x(x_1 - 2) + y(y_1 - 3) - 2x_1 - 3y_1 + 4 = 0\).
This polar equation must represent the given line \(x + 2by - 5 = 0\).
Comparing the coefficients of \(x\), \(y\), and the constant term:
\[ \frac{x_1 - 2}{1} = \frac{y_1 - 3}{2b} = \frac{-2x_1 - 3y_1 + 4}{-5} \] From the first two parts of the equality, let \(k\) be the common ratio: \(x_1 - 2 = k \cdot 1 \implies x_1 = k+2\)
\(y_1 - 3 = k \cdot 2b \implies y_1 = 2bk+3\)
Substitute these expressions for \(x_1\) and \(y_1\) into the third part of the equality:
\[ \frac{-2(k+2) - 3(2bk+3) + 4}{-5} = k \] \[ -2k - 4 - 6bk - 9 + 4 = -5k \] \[ -2k - 6bk - 9 = -5k \] Move terms involving \(k\) to one side and constants to the other: \[ -6bk + 3k = 9 \] \[ k(-6b + 3) = 9 \] \[ k(3 - 6b) = 9 \] \[ k = \frac{9}{3 - 6b} = \frac{3}{1 - 2b}. \] Now find the coordinates of the pole \((x_1, y_1)\): \[ x_1 = k+2 = \frac{3}{1-2b} + 2 = \frac{3 + 2(1-2b)}{1-2b} = \frac{3 + 2 - 4b}{1-2b} = \frac{5 - 4b}{1-2b} \] \[ y_1 = 2bk+3 = 2b\left(\frac{3}{1-2b}\right) + 3 = \frac{6b}{1-2b} + \frac{3(1-2b)}{1-2b} = \frac{6b + 3 - 6b}{1-2b} = \frac{3}{1-2b}. \] So, the pole is \(\left(\frac{5 - 4b}{1-2b}, \frac{3}{1-2b}\right)\). 
Step 2: Use the condition that the pole lies on the line \(x + by + 1 = 0\).
Substitute the coordinates of the pole \((x_1, y_1)\) into the equation of the line \(x + by + 1 = 0\): \[ \frac{5 - 4b}{1-2b} + b\left(\frac{3}{1-2b}\right) + 1 = 0 \] Multiply the entire equation by \((1-2b)\) to clear the denominator (assuming \(1-2b \neq 0\)): \[ (5 - 4b) + 3b + 1(1-2b) = 0 \] \[ 5 - 4b + 3b + 1 - 2b = 0 \] Combine the constant terms and the terms involving \(b\): \[ (5 + 1) + (-4b + 3b - 2b) = 0 \] \[ 6 - 3b = 0 \] \[ 3b = 6 \] \[ b = 2. \] Step 3: Find the polar of the point \((b, -b)\) with respect to the circle \(S = 0\).
We found that \(b=2\). So, the point is \((2, -2)\).
The circle is \(S = x^2 + y^2 - 4x - 6y + 4 = 0\), with \(g=-2, f=-3, c=4\).
The polar of a point \((x_1, y_1)\) with respect to \(S = 0\) is \(xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0\).
Substitute \(x_1=2, y_1=-2, g=-2, f=-3, c=4\): \[ x(2) + y(-2) - 2(x+2) - 3(y-2) + 4 = 0 \] \[ 2x - 2y - 2x - 4 - 3y + 6 + 4 = 0 \] Combine like terms: \[ (2x - 2x) + (-2y - 3y) + (-4 + 6 + 4) = 0 \] \[ 0x - 5y + 6 = 0 \] \[ -5y + 6 = 0 \] Multiplying by -1, we get: \[ 5y - 6 = 0. \] This matches Option (1).