One end of the diameter of the circle x^2+y^2 | vedclass

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(C) The given circle equation is $x^2+y^2-6x-5y-1=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3$ and $f=-2.5$. The center of the circle i

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$8x-y-54=0$

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(C) The given circle equation is $x^2+y^2-6x-5y-1=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3$ and $f=-2.5$. The center of the circle is $(-g, -f) = (3, 2.5)$. Let the other end of the diameter be $(x_2, y_2)$. Since the center $(3, 2.5)$ is the midpoint of the diameter with endpoints $(-1, 3)$ and $(x_2, y_2)$,we have: $\frac{-1+x_2}{2} = 3 \Rightarrow x_2 = 7$ $\frac{3+y_2}{2} = 2.5 \Rightarrow y_2 = 2$ So,the other end is $(7, 2)$. The equation of the tangent to the circle $x^2+y^2+2gx+2fy+c=0$ at point $(x_1, y_1)$ is given by $xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0$. Substituting $(x_1, y_1) = (7, 2)$,$g=-3$,$f=-2.5$,and $c=-1$: $7x + 2y - 3(x+7) - 2.5(y+2) - 1 = 0$ $7x + 2y - 3x - 21 - 2.5y - 5 - 1 = 0$ $4x - 0.5y - 27 = 0$ Multiplying by $2$,we get $8x - y - 54 = 0$.

One end of the diameter of the circle $x^2+y^2-6x-5y-1=0$ is $(-1,3)$. Find the equation of the tangent at the other end of the diameter.

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