If 2x - 3y + 1 = 0 is the equation of the pol | vedclass

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(C) The equation of the polar of a point $P(x_1, y_1)$ with respect to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $xx_1 + yy_1 + g(x + x_1

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$3$

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(C) The equation of the polar of a point $P(x_1, y_1)$ with respect to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0$.Given the circle $x^2 + y^2 - 2x + 4y + 3 = 0$,we have $g = -1$,$f = 2$,and $c = 3$.Substituting these values,the equation of the polar is $xx_1 + yy_1 - 1(x + x_1) + 2(y + y_1) + 3 = 0$.Rearranging the terms,we get $x(x_1 - 1) + y(y_1 + 2) - x_1 + 2y_1 + 3 = 0$.Comparing this with the given polar equation $2x - 3y + 1 = 0$,we can write the ratios of the coefficients:$\frac{x_1 - 1}{2} = \frac{y_1 + 2}{-3} = \frac{-x_1 + 2y_1 + 3}{1} = k$ (say).From the first two ratios,$x_1 - 1 = 2k \implies x_1 = 2k + 1$ and $y_1 + 2 = -3k \implies y_1 = -3k - 2$.Substituting these into the third ratio: $- (2k + 1) + 2(-3k - 2) + 3 = k$.$-2k - 1 - 6k - 4 + 3 = k \implies -8k - 2 = k \implies 9k = -2 \implies k = -2/9$.Now,$x_1 = 2(-2/9) + 1 = -4/9 + 9/9 = 5/9$ and $y_1 = -3(-2/9) - 2 = 6/9 - 18/9 = -12/9 = -4/3$.Finally,$3x_1 - y_1 = 3(5/9) - (-4/3) = 5/3 + 4/3 = 9/3 = 3$.

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 $3x_1 - y_1 =$

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