The line 3x-y+k=0 touches the circle x^2+y^2+ | vedclass

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(C) The line $3x-y+k=0$ touches the circle $x^2+y^2+4x-6y+3=0$. The center of the circle is $(-2, 3)$ and the radius $r = \sqrt{(-2)^2 + 3^2 - 3} = \s

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$x+16y-56=0$

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(C) The line $3x-y+k=0$ touches the circle $x^2+y^2+4x-6y+3=0$. The center of the circle is $(-2, 3)$ and the radius $r = \sqrt{(-2)^2 + 3^2 - 3} = \sqrt{4+9-3} = \sqrt{10}$. Since the line is tangent to the circle,the perpendicular distance from the center $(-2, 3)$ to the line $3x-y+k=0$ must be equal to the radius $r$. $\frac{|3(-2) - (3) + k|}{\sqrt{3^2 + (-1)^2}} = \sqrt{10}$ $|k-9| = \sqrt{10} 	imes \sqrt{10} = 10$ $k-9 = 10$ or $k-9 = -10$ $k = 19$ or $k = -1$. Given $k_1 The point is $(k_1, k_2) = (-1, 19)$. The equation of the chord of contact of a point $(x_1, y_1)$ with respect to the circle $x^2+y^2+2gx+2fy+c=0$ is $xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0$. Here,$g=2, f=-3, c=3, x_1=-1, y_1=19$. $-x + 19y + 2(x-1) - 3(y+19) + 3 = 0$ $-x + 19y + 2x - 2 - 3y - 57 + 3 = 0$ $x + 16y - 56 = 0$.

The line $3x-y+k=0$ touches the circle $x^2+y^2+4x-6y+3=0$. If $k_1, k_2$ $(k_1 < k_2)$ are the two values of $k$,then the equation of the chord of contact of the point $(k_1, k_2)$ with respect to the given circle is

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