Different approach to finding possible values of slope of line intersecting a given circle

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Given an equation of circle $(x-3)^2+(y+2)^2=25$ . A line $y=mx+1$ intesects the circle at points $P$ and $Q$ such that $x$ coordinate of the midpoint of $PQ$ is $\frac{-3}{5}.$

We have to find range of $m.$
Given answer is $m \in[2,4).$

My approach:

I did find out the $y$ coordinate by putting $x$ is given equation of line. Then since the line $PQ$ will be perpendicular to the line joining the point $(3,-2)$ and midpoint of line segment $PQ.$
Finally I equated the product of two slopes to $-1$ and hence solved the quadratic obtained in $m$ and got the possible values of $m.$

WHAT I WANT TO ASK:

Since the line intersects the circle in two points, I substituted $y=mx+1$ in circle and got quadratic equation in $x$ with coffiecients in terms of $m.$
Since I know that the line intersects in two real points I put the condition that discriminant of the quadratic obtained is greater than $0,$ but the quadratic obtained has non-real roots.
I want to ask what is wrong in the second approach ?

3

There are 3 best solutions below

1
On

The lines you are considering pass through the point $(0,1)$ which is an interior point for the circle. So every line intersects the circle in two distinct points.

It seems that your first line of reasoning is not quite correct, but the second is.

3
On

The second approach will give
$$x^2(m^2+1)+x(6m-6)-7=0.$$ Discriminant $D=64m^2-72m+64$ is positive for all $m\in \mathbb{R}.$

This signifies that for each $m,$ the line will cut the circle at two distinct points.
It is logical because all the lines pass through the point $(0,1)$ which lies inside the circle.

Let us continue.
The midpoint is arithmetic mean of $P$ and $Q.$ For $x$-coordinates that we know, we solve $$\frac{-6m+6}{2(m^2+1)}=-\frac{3}{5}$$ or equivalently $$m^2-5m+6=0.$$

(Is the given answer correct?)

0
On

Just to verify the answer given by @user376343 is correct. There will be only two discrete values of $m,\ 2\ \text{or}\ 3$ rather than a range of values. enter image description here