Question

If $2x - 3y + 1 = 0$ is the equation of the polar of a point $P(x_1, y_1)$ with respect to the circle $x^2 + y^2 - 2x + 4y + 3 = 0$, then $3x_1 - y_1 =$

• A. $\frac{1}{3}$
• B. $-3$
• C. $3$
• D. $-\frac{1}{3}$

Hint:

The polar of $(x_1, y_1)$ for a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0$. Compare coefficients to find $x_1, y_1$.

Answer 1

The Correct Option is C

The circle’s equation is $x^2 + y^2 - 2x + 4y + 3 = 0$, with $g = -1$, $f = 2$, $c = 3$. The polar of point $P(x_1, y_1)$ is: \[ xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0 \] \[ xx_1 + yy_1 - (x + x_1) + 2(y + y_1) + 3 = 0 \] \[ x(x_1 - 1) + y(y_1 + 2) + (-x_1 + 2y_1 + 3) = 0 \] Given the polar is $2x - 3y + 1 = 0$, compare coefficients: \[ x_1 - 1 = 2, \quad y_1 + 2 = -3, \quad -x_1 + 2y_1 + 3 = 1 \] From the first: \[ x_1 = 3 \] From the second: \[ y_1 = -5 \] Verify the third: \[ -3 + 2(-5) + 3 = -3 - 10 + 3 = -10 \neq 1 \] The third equation is inconsistent, suggesting a possible error. Solve using the first two: \[ 3x_1 - y_1 = 3 \cdot 3 - (-5) = 9 + 5 = 14 \neq 3 \] Recheck the polar equation derivation. The correct polar should yield consistent equations. Assume the given polar $2x - 3y + 1 = 0$ is correct and recompute using the ratio method: \[ \frac{x_1 - 1}{2} = \frac{y_1 + 2}{-3} = \frac{-x_1 + 2y_1 + 3}{1} = k \] \[ x_1 - 1 = 2k \implies x_1 = 2k + 1 \] \[ y_1 + 2 = -3k \implies y_1 = -3k - 2 \] \[ -x_1 + 2y_1 + 3 = k \implies -(2k + 1) + 2(-3k - 2) + 3 = k \] \[ -2k - 1 - 6k - 4 + 3 = k \implies -8k - 2 = k \implies -9k = 2 \implies k = -\frac{2}{9} \] \[ x_1 = 2 \left( -\frac{2}{9} \right) + 1 = -\frac{4}{9} + 1 = \frac{5}{9} \] \[ y_1 = -3 \left( -\frac{2}{9} \right) - 2 = \frac{6}{9} - 2 = \frac{2}{3} - \frac{6}{3} = -\frac{4}{3} \] \[ 3x_1 - y_1 = 3 \cdot \frac{5}{9} - \left( -\frac{4}{3} \right) = \frac{15}{9} + \frac{4}{3} = \frac{5}{3} + \frac{4}{3} = 3 \] Option (3) is correct. Options (1), (2), and (4) do not match.