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

Equation of line $PQ: y = mx + 1$ Let $R$ be the mid point of $PQ$, so its $x$-coordinate of midpoint $= - \dfrac{3}{5}$ Hence, coordinates of midpoint $R\left(- \dfrac{3}{5}, - \dfrac{3m}{5} + 1\right)$ Let centre of circle is $O\left(3, 2\right)$ Since, perpendicular dropped from centre upon any chord divides it in two equal parts. Hence, line $PQ$ will be perpendicular to $RO$ $⇒ {m}_{RQ} ⋅ {m}_{RO} = - 1$ $⇒ m\left(\dfrac{- \dfrac{3m}{5} + 1 + 2}{- \dfrac{3}{5} - 3}\right) = - 1$ $⇒ m\left(\dfrac{- 3m + 15}{- 18}\right) = - 1$ $⇒ {m}^{2} - 5m + 6 = 0$ $⇒ m = 2, 3$

$⇒ 2 \le m < 4$