A line y=mx+1 intersects the circle (x-3)^2+( | vedclass

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(B) The circle is $(x-3)^2+(y+2)^2=25$,so its center is $C(3, -2)$.Let $R$ be the midpoint of the chord $PQ$. Since $R$ lies on the line $y=mx+1$,its

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$2 \leq m < 4$

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(B) The circle is $(x-3)^2+(y+2)^2=25$,so its center is $C(3, -2)$.Let $R$ be the midpoint of the chord $PQ$. Since $R$ lies on the line $y=mx+1$,its coordinates are $(x_R, mx_R+1)$. Given $x_R = -\frac{3}{5}$,we have $y_R = m(-\frac{3}{5}) + 1 = \frac{-3m+5}{5}$.Thus,$R = (-\frac{3}{5}, \frac{-3m+5}{5})$.The line segment $CR$ is perpendicular to the chord $PQ$. The slope of $PQ$ is $m$,so the slope of $CR$ must be $-\frac{1}{m}$.Slope of $CR = \frac{y_R - (-2)}{x_R - 3} = \frac{\frac{-3m+5}{5} + 2}{-\frac{3}{5} - 3} = \frac{-3m+5+10}{-3-15} = \frac{-3m+15}{-18} = \frac{m-5}{6}$.Equating the slopes: $\frac{m-5}{6} = -\frac{1}{m}$.$m(m-5) = -6 \Rightarrow m^2 - 5m + 6 = 0$.$(m-2)(m-3) = 0$,so $m=2$ or $m=3$.Both values $m=2$ and $m=3$ satisfy the condition $2 \leq m < 4$.

$A$ line $y=mx+1$ intersects the circle $(x-3)^2+(y+2)^2=25$ at the points $P$ and $Q$. If the midpoint of the line segment $PQ$ has $x$-coordinate $-\frac{3}{5}$,then which one of the following options is correct?

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