The equations of the normals to the circle x^ | vedclass

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(A) The given circle equation is $x^2 + y^2 - 8x - 2y + 12 = 0$. The center of the circle is $(4, 1)$.Substitute $y = -1$ into the circle equation to

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$2x - y - 7 = 0, 2x + y - 9 = 0$

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(A) The given circle equation is $x^2 + y^2 - 8x - 2y + 12 = 0$. The center of the circle is $(4, 1)$.Substitute $y = -1$ into the circle equation to find the abscissa:$x^2 + (-1)^2 - 8x - 2(-1) + 12 = 0$$x^2 - 8x + 1 + 2 + 12 = 0$$x^2 - 8x + 15 = 0$$(x - 3)(x - 5) = 0$So,$x = 3$ or $x = 5$. The points are $(3, -1)$ and $(5, -1)$.The normal to a circle at any point always passes through the center $(4, 1)$.For point $(3, -1)$,the slope of the normal is $m = \frac{1 - (-1)}{4 - 3} = \frac{2}{1} = 2$.The equation is $y - (-1) = 2(x - 3)$ $\Rightarrow y + 1 = 2x - 6$ $\Rightarrow 2x - y - 7 = 0$.For point $(5, -1)$,the slope of the normal is $m = \frac{1 - (-1)}{4 - 5} = \frac{2}{-1} = -2$.The equation is $y - (-1) = -2(x - 5)$ $\Rightarrow y + 1 = -2x + 10$ $\Rightarrow 2x + y - 9 = 0$.

The equations of the normals to the circle $x^2 + y^2 - 8x - 2y + 12 = 0$ at the points whose ordinate is $-1$ are:

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