If (4,2) and (k,-3) are conjugate points with | vedclass

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

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$\frac{28}{3}$

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(A) The equation of the polar of point $(x_1, y_1)$ with respect to the circle $x^2+y^2+2gx+2fy+c=0$ is given by $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$. For the circle $x^2+y^2-5x+8y+6=0$,we have $g = -\frac{5}{2}$,$f = 4$,and $c = 6$. The polar of point $(4,2)$ is: $x(4) + y(2) - \frac{5}{2}(x+4) + 4(y+2) + 6 = 0$ $4x + 2y - \frac{5}{2}x - 10 + 4y + 8 + 6 = 0$ Multiplying by $2$: $8x + 4y - 5x - 20 + 8y + 16 + 12 = 0$ $3x + 12y + 8 = 0$ Since $(k,-3)$ is a conjugate point,it must lie on the polar of $(4,2)$. Substituting $(k,-3)$ into the polar equation: $3(k) + 12(-3) + 8 = 0$ $3k - 36 + 8 = 0$ $3k - 28 = 0$ $k = \frac{28}{3}$

If $(4,2)$ and $(k,-3)$ are conjugate points with respect to $x^2+y^2-5x+8y+6=0$,then $k$ equals

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