A line y=mx+1 meets the circle `(x-3)^(2)+(y+2)^(2)=25` at point P and Q. if mid point of PQ has abscissa of `-(3)/(5)` then value of m satisfies

Options

A. `4 le mlt 6`
B. ` -3 le m lt - 1`
C. ` 2 le m lt 4`
D. `6le m lt 8`

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the equations of the circle and the line
The equation of the circle is given as:
\[ (x - 3)^2 + (y + 2)^2 = 25 \]
This represents a circle with center \(C(3, -2)\) and radius \(r = 5\).

The equation of the line is given as:
\[ y = mx + 1 \]

### Step 2: Find the midpoint of the points of intersection
Let the points of intersection of the line and the circle be \(P\) and \(Q\). The midpoint \(M\) of \(PQ\) has an abscissa (x-coordinate) of \(-\frac{3}{5}\).

### Step 3: Substitute the x-coordinate into the line equation
Substituting \(x = -\frac{3}{5}\) into the line equation to find the y-coordinate:
\[ y = m\left(-\frac{3}{5}\right) + 1 = 1 - \frac{3m}{5} \]
Thus, the coordinates of the midpoint \(M\) are:
\[ M\left(-\frac{3}{5}, 1 - \frac{3m}{5}\right) \]

### Step 4: Determine the slope of line segment \(CR\) (where \(R\) is the midpoint)
The slope of the line \(CR\) (from center \(C(3, -2)\) to midpoint \(M\)) is given by:
\[ \text{slope of } CR = \frac{y_M - y_C}{x_M - x_C} = \frac{\left(1 - \frac{3m}{5}\right) - (-2)}{-\frac{3}{5} - 3} \]
Calculating this gives:
\[ = \frac{1 - \frac{3m}{5} + 2}{-\frac{3}{5} - \frac{15}{5}} = \frac{3 - \frac{3m}{5}}{-\frac{18}{5}} = \frac{3 - \frac{3m}{5}}{-\frac{18}{5}} = \frac{5(3 - \frac{3m}{5})}{-18} \]
Simplifying further:
\[ = \frac{15 - 3m}{-18} = \frac{3m - 15}{18} \]

### Step 5: Set up the relationship between slopes
Since \(CR\) is perpendicular to \(PQ\), the product of their slopes must equal \(-1\). The slope of line \(y = mx + 1\) is \(m\). Therefore:
\[ m \cdot \frac{3m - 15}{18} = -1 \]
This simplifies to:
\[ 3m^2 - 15m + 18 = 0 \]

### Step 6: Solve the quadratic equation
Dividing the entire equation by 3:
\[ m^2 - 5m + 6 = 0 \]
Factoring gives:
\[ (m - 2)(m - 3) = 0 \]
Thus, the solutions for \(m\) are:
\[ m = 2 \quad \text{or} \quad m = 3 \]

### Step 7: Conclusion
The values of \(m\) that satisfy the condition are \(m = 2\) and \(m = 3\).