Question

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

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

Answer 1

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

Centre of circle $C\equiv (3,-2)$
Given $x=\frac{-3}{5}$ lies on $y=mx+1$

$y=1-\frac{3m}{5}\therefore R=(\frac{-3}{5},1-\frac{3m}{5})$

Slope $m_{CR}=\frac{1-\frac{3m}{5}+2}{\frac{-3}{5}-3}=\frac{\frac{15-3m}{5}}{\frac{-3-15}{5}}=\frac{15-3m}{-18}=\frac{m-5}{6}$

$CR\perp PQ\Rightarrow \frac{m-5}{6}\times m=-1\Rightarrow m^2-5m+6=0\Rightarrow (m-2)(m-3)=0\therefore m=2,3\Rightarrow 2\le m<4$