The mid-point of the chord of the circle x^2+ | vedclass

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(C) The equation of the circle is $x^2+y^2-6x+4y-12=0$. The center $C$ is $(3, -2)$ and the radius $r = \sqrt{3^2 + (-2)^2 - (-12)} = \sqrt{9+4+12} =

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$\left(-\frac{1}{5}, \frac{2}{5}\right)$

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(C) The equation of the circle is $x^2+y^2-6x+4y-12=0$. The center $C$ is $(3, -2)$ and the radius $r = \sqrt{3^2 + (-2)^2 - (-12)} = \sqrt{9+4+12} = 5$.The slope of the tangent at $(-1, 1)$ is found by differentiating the circle equation: $2x + 2yy' - 6 + 4y' = 0 \Rightarrow y' = \frac{6-2x}{2y+4}$. At $(-1, 1)$,$y' = \frac{6-2(-1)}{2(1)+4} = \frac{8}{6} = \frac{4}{3}$.The equation of the tangent is $y-1 = \frac{4}{3}(x+1) \Rightarrow 4x-3y+7=0$.$A$ chord parallel to the tangent has the form $4x-3y+k=0$. The distance between the tangent and the chord is $\frac{|k-7|}{\sqrt{4^2+(-3)^2}} = 1$ $\Rightarrow |k-7| = 5$ $\Rightarrow k = 12$ or $k = 2$.The distance of the line $4x-3y+k=0$ from the center $(3, -2)$ is $d = \frac{|4(3)-3(-2)+k|}{5} = \frac{|12+6+k|}{5} = \frac{|18+k|}{5}$.For $k=12$,$d = \frac{30}{5} = 6 > r$ (no chord).For $k=2$,$d = \frac{20}{5} = 4 The chord is $4x-3y+2=0$. The line $CP$ is perpendicular to the chord,so its slope is $-\frac{3}{4}$ and it passes through $(3, -2)$.Equation of $CP$: $y+2 = -\frac{3}{4}(x-3) \Rightarrow 3x+4y=1$.Solving $4x-3y+2=0$ and $3x+4y=1$ gives $x = -\frac{1}{5}, y = \frac{2}{5}$.

The mid-point of the chord of the circle $x^2+y^2-6x+4y-12=0$ drawn parallel to the tangent at $(-1,1)$ and at a distance of one unit from the tangent is

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