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

Answer 1

The correct option is B

$2\le m<4$

The explanation for the correct option:

The given line $\begin{array}{rcl}y& =& mx+1\end{array}$intersects ${(x-3)}^2+{(y+2)}^2=25$

$\begin{array}{rcl}{(x–3)}^2+({(mx+1)+2)}^2& =& 25\\ x^2(1+m^2)+6(m-1)x–7& =& 0\\ \frac{-3}{5}& =& \frac{\alpha +\beta }{2}=\frac{-6(m-1)}{2(1+m^2)}\\ 1+m^2& =& 5m–5\\ m^2-5m+5+1& =& 0\\ m^2-5m+6& =& 0\\ m& =& 2,3\end{array}$

Hence, the correct answer is option (B)