One end of a diameter of a circle
`x^(2) + y^(2) - 3x + 5y - 4 = 0 `
is (1, - 6) , find the other end .

Answer

To find the other end of the diameter of the circle given by the equation \( x^2 + y^2 - 3x + 5y - 4 = 0 \) with one end at the point \( (1, -6) \), we will follow these steps:

### Step 1: Rewrite the Circle Equation
First, we will rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 - 3x + 5y - 4 = 0 \]
We will complete the square for both \( x \) and \( y \).

### Step 2: Completing the Square
For \( x \):
1. Take the coefficient of \( x \) which is \(-3\), halve it to get \(-\frac{3}{2}\), and square it to get \(\left(-\frac{3}{2}\right)^2 = \frac{9}{4}\).
2. For \( y \):
- Take the coefficient of \( y \) which is \(5\), halve it to get \(\frac{5}{2}\), and square it to get \(\left(\frac{5}{2}\right)^2 = \frac{25}{4}\).

Now we can rewrite the equation:
\[ (x^2 - 3x + \frac{9}{4}) + (y^2 + 5y + \frac{25}{4}) = 4 + \frac{9}{4} + \frac{25}{4} \]
This simplifies to:
\[ (x - \frac{3}{2})^2 + (y + \frac{5}{2})^2 = \frac{34}{4} = \frac{17}{2} \]

### Step 3: Identify the Center and Radius
From the standard form of the circle \((x - h)^2 + (y - k)^2 = r^2\), we identify:
- Center \( C(h, k) = \left(\frac{3}{2}, -\frac{5}{2}\right) \)
- Radius \( r = \sqrt{\frac{17}{2}} \)

### Step 4: Use the Midpoint Formula
Let the other end of the diameter be \( A(h, k) \). The midpoint \( M \) of the diameter is the center of the circle:
\[ M = \left(\frac{1 + h}{2}, \frac{-6 + k}{2}\right) \]
Setting this equal to the center \( \left(\frac{3}{2}, -\frac{5}{2}\right) \):
\[ \frac{1 + h}{2} = \frac{3}{2} \quad \text{and} \quad \frac{-6 + k}{2} = -\frac{5}{2} \]

### Step 5: Solve for \( h \) and \( k \)
From the first equation:
\[ 1 + h = 3 \implies h = 2 \]
From the second equation:
\[ -6 + k = -5 \implies k = 1 \]

### Conclusion
The coordinates of the other end of the diameter are \( (h, k) = (2, 1) \).

### Final Answer
The other end of the diameter is \( (2, 1) \).

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