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 ?

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 -3/5, then which one of the following options is correct ?

$\text{Given: } y = mx + 1 \text{ cuts } (x-3)^2 + (y+2)^2 = 25 \text{ at } P,\ Q$

$\text{Midpoint of } PQ \text{ has } x = -\dfrac{3}{5}$

$\text{To find: which range of } m \text{ is correct}$

$\text{Circle}: (x-3)^2 + (y+2)^2 = 25 \Rightarrow C(3,-2),\ r = 5$

$y = mx + 1 \text{ always passes through } A(0, 1)$

$M = \text{foot of perpendicular from } C \text{ onto the line}$

$\text{Line: } mx - y + 1 = 0 \ (a = m,\ b = -1,\ c = 1)$

$x_M = x_C - a\cdot\dfrac{a x_C + b y_C + c}{a^2 + b^2}$

$x_M = 3 - \dfrac{m(3m + 3)}{m^2 + 1}$

$3 - \dfrac{m(3m+3)}{m^2+1} = -\dfrac{3}{5}$

$\Rightarrow \dfrac{m(3m+3)}{m^2+1} = \dfrac{18}{5}$

$5(3m^2 + 3m) = 18(m^2 + 1)$

$3m^2 - 15m + 18 = 0$

$m^2 - 5m + 6 = 0$

$(m-2)(m-3) = 0 \Rightarrow m = 2 \text{ or } m = 3$

$d = \dfrac{|3m + 3|}{\sqrt{m^2 + 1}}$

$m = 2:\ \dfrac{9}{\sqrt5} < 5, \qquad m = 3:\ \dfrac{12}{\sqrt{10}} < 5$

∠ CMA = 90^∘ ⇒ M lies on the circle with diameter CA

$m \in \{2,\ 3\} \subset [2,\ 4)$

$\boxed{\text{option (1): } 2 \le m < 4}$