Question

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. $-3\le m<-1$
• B. $2\le m<4$
• C. $4\le m<6$
• D. $6\le m<8$

Answer 1

The correct option is B $2\le m<4$


Coordinate of midpoint
$R=(-\frac{3}{5},-\frac{3}{5}m+1)(Slope{)}_{PQ}\times (Slope{)}_{CR}=-1\Rightarrow m\times \frac{-\frac{3}{5}m+1+2}{-\frac{3}{5}-3}=-1\Rightarrow m\times \frac{m-5}{6}=-1\Rightarrow m^2-5m+6=0\Rightarrow (m-2)(m-3)=0\Rightarrow m=2,3\therefore 2\le m<4$