Show that the common chord of the circles `x^(2)+y^(2)-6x-4y+9=0andx^(2)+y^(2)-8x-6y+23=0` paas through the centre of the second circle and find its length.

Answer

To solve the problem step by step, we will first derive the equations of the circles, find the equation of the common chord, check if it passes through the center of the second circle, and finally calculate the length of the common chord.

### Step 1: Write the equations of the circles

The equations of the circles are given as:
1. Circle 1: \( x^2 + y^2 - 6x - 4y + 9 = 0 \)
2. Circle 2: \( x^2 + y^2 - 8x - 6y + 23 = 0 \)

### Step 2: Find the centers and radii of the circles

To find the center and radius of each circle, we can rewrite the equations in standard form.

**For Circle 1:**
- Rearranging the equation:
\[ x^2 - 6x + y^2 - 4y + 9 = 0 \]
- Completing the square:
\[ (x^2 - 6x + 9) + (y^2 - 4y + 4) = 4 \]
\[ (x - 3)^2 + (y - 2)^2 = 4 \]
- Center: \( (3, 2) \), Radius: \( r_1 = 2 \)

**For Circle 2:**
- Rearranging the equation:
\[ x^2 - 8x + y^2 - 6y + 23 = 0 \]
- Completing the square:
\[ (x^2 - 8x + 16) + (y^2 - 6y + 9) = 2 \]
\[ (x - 4)^2 + (y - 3)^2 = 2 \]
- Center: \( (4, 3) \), Radius: \( r_2 = \sqrt{2} \)

### Step 3: Find the equation of the common chord

The equation of the common chord can be found using the formula \( S_1 - S_2 = 0 \), where \( S_1 \) and \( S_2 \) are the left-hand sides of the circle equations.

Calculating \( S_1 - S_2 \):
\[ S_1 = x^2 + y^2 - 6x - 4y + 9 \]
\[ S_2 = x^2 + y^2 - 8x - 6y + 23 \]
Subtracting these:
\[ S_1 - S_2 = (-6x + 8x) + (-4y + 6y) + (9 - 23) = 2x + 2y - 14 = 0 \]
Simplifying gives:
\[ x + y - 7 = 0 \]
Thus, the equation of the common chord is \( x + y = 7 \).

### Step 4: Check if the common chord passes through the center of the second circle

The center of the second circle is \( (4, 3) \). We substitute these values into the equation of the common chord:
\[ 4 + 3 - 7 = 0 \]
Since this is true, the common chord does indeed pass through the center of the second circle.

### Step 5: Find the length of the common chord

The common chord is a diameter of the second circle. The length of the diameter is \( 2r \).

We already found that the radius \( r_2 \) of the second circle is \( \sqrt{2} \). Therefore, the length of the common chord (diameter) is:
\[ \text{Length} = 2r_2 = 2 \times \sqrt{2} = 2\sqrt{2} \]

### Final Answer
The common chord passes through the center of the second circle, and its length is \( 2\sqrt{2} \).

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