If x+y=3 is the equation of the chord AB of the circle `x^2+y^2-2x+4y-8=0`, find the equation of the circle having as diameter.

Answer

To solve the problem, we need to find the equation of the circle that has the chord AB defined by the line \(x + y = 3\) as its diameter, given that the chord lies on the circle defined by the equation \(x^2 + y^2 - 2x + 4y - 8 = 0\).

### Step-by-Step Solution:

1. **Identify the Circle's Equation**:
The given equation of the circle is:
\[ x^2 + y^2 - 2x + 4y - 8 = 0 \]
We can rewrite it in standard form by completing the square.

2. **Complete the Square**:
Rearranging the equation:
\[ (x^2 - 2x) + (y^2 + 4y) = 8 \]
Completing the square for \(x\):
\[ x^2 - 2x = (x - 1)^2 - 1 \]
Completing the square for \(y\):
\[ y^2 + 4y = (y + 2)^2 - 4 \]
Substituting back:
\[ (x - 1)^2 - 1 + (y + 2)^2 - 4 = 8 \]
Simplifying:
\[ (x - 1)^2 + (y + 2)^2 = 13 \]
This shows that the circle is centered at \((1, -2)\) with a radius of \(\sqrt{13}\).

3. **Find the Points of Intersection**:
The chord \(AB\) is given by the line \(x + y = 3\). We can express \(y\) in terms of \(x\):
\[ y = 3 - x \]
Substitute \(y\) into the circle's equation:
\[ x^2 + (3 - x)^2 - 2x + 4(3 - x) - 8 = 0 \]
Expanding:
\[ x^2 + (9 - 6x + x^2) - 2x + 12 - 4x - 8 = 0 \]
Combine like terms:
\[ 2x^2 - 12x + 13 = 0 \]

4. **Solve the Quadratic Equation**:
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ x = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 2 \cdot 13}}{2 \cdot 2} \]
Simplifying:
\[ x = \frac{12 \pm \sqrt{144 - 104}}{4} = \frac{12 \pm \sqrt{40}}{4} = \frac{12 \pm 2\sqrt{10}}{4} = 3 \pm \frac{\sqrt{10}}{2} \]
Thus, the \(x\)-coordinates of points A and B are:
\[ x_1 = 3 + \frac{\sqrt{10}}{2}, \quad x_2 = 3 - \frac{\sqrt{10}}{2} \]

5. **Find Corresponding \(y\)-Coordinates**:
Using \(y = 3 - x\):
\[ y_1 = 3 - x_1 = 3 - \left(3 + \frac{\sqrt{10}}{2}\right) = -\frac{\sqrt{10}}{2} \]
\[ y_2 = 3 - x_2 = 3 - \left(3 - \frac{\sqrt{10}}{2}\right) = \frac{\sqrt{10}}{2} \]

6. **Find the Center of the Circle with Diameter AB**:
The center \((h, k)\) of the circle with diameter AB is:
\[ h = \frac{x_1 + x_2}{2} = \frac{(3 + \frac{\sqrt{10}}{2}) + (3 - \frac{\sqrt{10}}{2})}{2} = \frac{6}{2} = 3 \]
\[ k = \frac{y_1 + y_2}{2} = \frac{-\frac{\sqrt{10}}{2} + \frac{\sqrt{10}}{2}}{2} = 0 \]

7. **Find the Radius**:
The length of diameter AB is:
\[ d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} = \sqrt{\left(\frac{\sqrt{10}}{2} + \frac{\sqrt{10}}{2}\right)^2 + \left(-\frac{\sqrt{10}}{2} - \frac{\sqrt{10}}{2}\right)^2} \]
Simplifying:
\[ d = \sqrt{10 + 10} = \sqrt{20} = 2\sqrt{5} \]
Thus, the radius \(r\) is:
\[ r = \frac{d}{2} = \sqrt{5} \]

8. **Equation of the Circle**:
The equation of the circle is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \(h = 3\), \(k = 0\), and \(r = \sqrt{5}\):
\[ (x - 3)^2 + (y - 0)^2 = 5 \]
Therefore, the required equation of the circle is:
\[ (x - 3)^2 + y^2 = 5 \]