For all real values of k,the point which lies | vedclass

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

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$(3, -1)$

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(A) The equation of the polar of a point $(x_1, y_1)$ with respect to the circle $x^2+y^2+2gx+2fy+c=0$ is given by $T=0$,which is $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$. Given the circle $x^2+y^2+4x-8y-5=0$,we have $g=2, f=-4, c=-5$. The polar of $(k, k+1)$ is: $kx + (k+1)y + 2(x+k) - 4(y+k+1) - 5 = 0$ $kx + ky + y + 2x + 2k - 4y - 4k - 4 - 5 = 0$ $(k+2)x + (k-3)y - 2k - 9 = 0$ Rearranging the terms to isolate $k$: $k(x+y-2) + (2x-3y-9) = 0$ For this to be true for all real values of $k$,both coefficients must be zero: $x+y-2 = 0$ and $2x-3y-9 = 0$. Solving these equations: From the first,$y = 2-x$. Substituting into the second: $2x - 3(2-x) - 9 = 0 \Rightarrow 2x - 6 + 3x - 9 = 0 \Rightarrow 5x = 15 \Rightarrow x = 3$. Then $y = 2-3 = -1$. The point is $(3, -1)$.

For all real values of $k$,the point which lies on the polar of $(k, k+1)$ with respect to the circle $x^2+y^2+4x-8y-5=0$ is

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