The point of concurrence of the polars of the | vedclass

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(D) 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 $xx_1 + yy_1 + g(x+x_1) + f(y+y_1)

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$(3,1)$

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(D) 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 $xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0$. Given the circle $x^2+y^2-4x-6y+1=0$,we have $g=-2, f=-3, c=1$. The polar of the point $(2t, t-4)$ is: $x(2t) + y(t-4) - 2(x+2t) - 3(y+t-4) + 1 = 0$ $2tx + ty - 4y - 2x - 4t - 3y - 3t + 12 + 1 = 0$ $2tx + ty - 2x - 7y - 7t + 13 = 0$ Rearranging the terms to group $t$: $t(2x + y - 7) + (-2x - 7y + 13) = 0$ For this to be concurrent for all $t \in R$,both parts must be zero: $2x + y - 7 = 0$ (Equation $1$) $-2x - 7y + 13 = 0$ (Equation $2$) Adding Equation $1$ and Equation $2$: $(2x - 2x) + (y - 7y) + (-7 + 13) = 0$ $-6y + 6 = 0 \implies y = 1$ Substituting $y=1$ into Equation $1$: $2x + 1 - 7 = 0 \implies 2x = 6 \implies x = 3$ The point of concurrence is $(3, 1)$.

The point of concurrence of the polars of the variable point $(2t, t-4)$,where $t \in R$,with respect to the circle $x^2+y^2-4x-6y+1=0$ is

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