Find the mid point of the chord intercepted by the circle `x^(2)+y^(2)-2x-10y+1=0` on the line `x-2y+7=0`

Answer

To find the midpoint of the chord intercepted by the circle \(x^2 + y^2 - 2x - 10y + 1 = 0\) on the line \(x - 2y + 7 = 0\), we can follow these steps:

### Step 1: Rewrite the Circle Equation
First, we rewrite the equation of the circle in standard form by completing the square.

The given equation is:
\[ x^2 + y^2 - 2x - 10y + 1 = 0 \]

Rearranging gives:
\[ x^2 - 2x + y^2 - 10y + 1 = 0 \]

Completing the square for \(x\):
\[ x^2 - 2x = (x - 1)^2 - 1 \]

Completing the square for \(y\):
\[ y^2 - 10y = (y - 5)^2 - 25 \]

Substituting back, we have:
\[ (x - 1)^2 - 1 + (y - 5)^2 - 25 + 1 = 0 \]
\[ (x - 1)^2 + (y - 5)^2 - 25 = 0 \]
\[ (x - 1)^2 + (y - 5)^2 = 25 \]

This shows that the circle has a center at \((1, 5)\) and a radius of \(5\).

### Step 2: Find the Equation of the Perpendicular Line
The line given is:
\[ x - 2y + 7 = 0 \]

We can rewrite it in slope-intercept form:
\[ y = \frac{1}{2}x + \frac{7}{2} \]

The slope of this line is \(\frac{1}{2}\). The slope of the line perpendicular to it is the negative reciprocal:
\[ -\frac{1}{\frac{1}{2}} = -2 \]

Using the point-slope form of the line, the equation of the line passing through the center \((1, 5)\) with slope \(-2\) is:
\[ y - 5 = -2(x - 1) \]
Expanding this gives:
\[ y - 5 = -2x + 2 \implies 2x + y - 7 = 0 \]

### Step 3: Solve the System of Equations
Now we have two equations:
1. \(x - 2y + 7 = 0\)
2. \(2x + y - 7 = 0\)

We can solve this system of equations. From the first equation, we can express \(x\) in terms of \(y\):
\[ x = 2y - 7 \]

Substituting this into the second equation:
\[ 2(2y - 7) + y - 7 = 0 \]
\[ 4y - 14 + y - 7 = 0 \]
\[ 5y - 21 = 0 \implies y = \frac{21}{5} \]

Now substituting \(y\) back to find \(x\):
\[ x = 2\left(\frac{21}{5}\right) - 7 = \frac{42}{5} - \frac{35}{5} = \frac{7}{5} \]

### Step 4: Conclusion
The midpoint \(P\) of the chord is:
\[ \left(\frac{7}{5}, \frac{21}{5}\right) \]

### Final Answer
The required midpoint of the chord is:
\[ \left(\frac{7}{5}, \frac{21}{5}\right) \]