Points P(-3, 2),Q(9, 10),and R(alpha, 4) lie | vedclass

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(A) Since $PR$ is the diameter,$\angle PQR = 90^\circ$. Thus,the product of the slopes of $PQ$ and $QR$ is $-1$.$m_{PQ} = \frac{10-2}{9-(-3)} = \frac{

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$3$

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(A) Since $PR$ is the diameter,$\angle PQR = 90^\circ$. Thus,the product of the slopes of $PQ$ and $QR$ is $-1$.$m_{PQ} = \frac{10-2}{9-(-3)} = \frac{8}{12} = \frac{2}{3}$.$m_{QR} = \frac{4-10}{\alpha-9} = \frac{-6}{\alpha-9}$.Since $m_{PQ} \cdot m_{QR} = -1$,we have $\frac{2}{3} \cdot \left(\frac{-6}{\alpha-9}\right) = -1$ $\Rightarrow \frac{-4}{\alpha-9} = -1$ $\Rightarrow \alpha-9 = 4$ $\Rightarrow \alpha = 13$.So,$R = (13, 4)$. The center $O$ of the circle is the midpoint of $PR$,$O = \left(\frac{-3+13}{2}, \frac{2+4}{2}\right) = (5, 3)$.The tangent at $Q(9, 10)$ is perpendicular to the radius $OQ$. $m_{OQ} = \frac{10-3}{9-5} = \frac{7}{4}$.Slope of tangent $QS = -\frac{4}{7}$.Equation of $QS$: $y-10 = -\frac{4}{7}(x-9)$ $\Rightarrow 7y - 70 = -4x + 36$ $\Rightarrow 4x + 7y = 106 \quad (1)$.The tangent at $R(13, 4)$ is perpendicular to the radius $OR$. $m_{OR} = \frac{4-3}{13-5} = \frac{1}{8}$.Slope of tangent $RS = -8$.Equation of $RS$: $y-4 = -8(x-13)$ $\Rightarrow y-4 = -8x + 104$ $\Rightarrow 8x + y = 108 \quad (2)$.Solving $(1)$ and $(2)$: From $(2)$,$y = 108 - 8x$. Substitute into $(1)$:$4x + 7(108 - 8x) = 106$ $\Rightarrow 4x + 756 - 56x = 106$ $\Rightarrow 52x = 650$ $\Rightarrow x = \frac{650}{52} = 12.5 = \frac{25}{2}$.$y = 108 - 8(\frac{25}{2}) = 108 - 100 = 8$.$S = (12.5, 8)$. Since $S$ lies on $2x - ky = 1$:$2(12.5) - k(8) = 1$ $\Rightarrow 25 - 8k = 1$ $\Rightarrow 8k = 24$ $\Rightarrow k = 3$.

Points $P(-3, 2)$,$Q(9, 10)$,and $R(\alpha, 4)$ lie on a circle $C$ with $PR$ as its diameter. The tangents to $C$ at the points $Q$ and $R$ intersect at the point $S$. If $S$ lies on the line $2x - ky = 1$,then $k$ is equal to $.........$.

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