If the pole of the line x+2by-5=0 with respec | vedclass

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(D) The equation of the circle is $S \equiv x^2+y^2-4x-6y+4=0$. The center is $(2,3)$ and the radius is $r = \sqrt{2^2+3^2-4} = 3$. Let the pole be $(

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$5x+y-6=0$

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(D) The equation of the circle is $S \equiv x^2+y^2-4x-6y+4=0$. The center is $(2,3)$ and the radius is $r = \sqrt{2^2+3^2-4} = 3$. Let the pole be $(x_1, y_1)$. The polar of $(x_1, y_1)$ with respect to $S=0$ is $xx_1 + yy_1 - 2(x+x_1) - 3(y+y_1) + 4 = 0$,which simplifies to $x(x_1-2) + y(y_1-3) - 2x_1 - 3y_1 + 4 = 0$. Comparing this with $x+2by-5=0$,we get $\frac{x_1-2}{1} = \frac{y_1-3}{2b} = \frac{-(2x_1+3y_1-4)}{-5} = k$. So,$x_1 = k+2$ and $y_1 = 2bk+3$. Since the pole $(x_1, y_1)$ lies on $x+by+1=0$,we have $(k+2) + b(2bk+3) + 1 = 0$,which gives $k(1+2b^2) + 3b+3 = 0$. Also,from the ratio,$5(x_1-2) = 2x_1+3y_1-4 \implies 3x_1-3y_1-6 = 0 \implies x_1-y_1-2=0$. Substituting $x_1, y_1$,we get $k+2 - (2bk+3) - 2 = 0 \implies k(1-2b) = 3 \implies k = \frac{3}{1-2b}$. Solving for $b$,we find $b=1$. For $b=1$,the point is $(1, -1)$. The polar of $(1, -1)$ is $x(1) + y(-1) - 2(x+1) - 3(y-1) + 4 = 0$,which simplifies to $x-y-2x-2-3y+3+4 = 0$,resulting in $-x-4y+5=0$ or $x+4y-5=0$. Re-evaluating the condition,if $b=1$,the polar of $(1, -1)$ is $x(1) + y(-1) - 2(x+1) - 3(y-1) + 4 = 0 \implies x-y-2x-2-3y+3+4 = 0 \implies -x-4y+5=0$. Given the options,the correct polar is $5x+y-6=0$.

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

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