To find the equation of the circle that passes through the points (2,0) and (0,2) and is orthogonal to the circle given by the equation \(2x^2 + 2y^2 + 5x - 6y + 4 = 0\), we can follow these steps:
### Step 1: General Equation of the Circle
Assume the equation of the required circle is of the form:
\[
x^2 + y^2 + 2gx + 2fy + c = 0
\]
where \(g\), \(f\), and \(c\) are constants we need to determine.
### Step 2: Substitute the Points into the Circle Equation
Since the circle passes through the points (2,0) and (0,2), we can substitute these points into the circle's equation.
**For point (2,0):**
\[
2^2 + 0^2 + 2g(2) + 2f(0) + c = 0
\]
This simplifies to:
\[
4 + 4g + c = 0 \quad \text{(Equation 1)}
\]
**For point (0,2):**
\[
0^2 + 2^2 + 2g(0) + 2f(2) + c = 0
\]
This simplifies to:
\[
4 + 4f + c = 0 \quad \text{(Equation 2)}
\]
### Step 3: Solve the Equations for \(g\) and \(f\)
From Equations 1 and 2, we have:
1. \(4 + 4g + c = 0\)
2. \(4 + 4f + c = 0\)
Setting both equations equal to each other (since they both equal \(-4\)):
\[
4g + c = 4f + c
\]
This leads to:
\[
4g = 4f \implies g = f
\]
### Step 4: Orthogonality Condition
The condition for orthogonality of two circles is given by:
\[
2g_1g_2 + 2f_1f_2 = c_1 + c_2
\]
where \(g_1, f_1, c_1\) are the coefficients of the first circle and \(g_2, f_2, c_2\) are the coefficients of the second circle.
For the given circle \(2x^2 + 2y^2 + 5x - 6y + 4 = 0\), we can rewrite it as:
\[
x^2 + y^2 + \frac{5}{2}x - 3y + 2 = 0
\]
Thus, we have:
- \(g_2 = \frac{5}{4}\)
- \(f_2 = -\frac{3}{2}\)
- \(c_2 = 2\)
Substituting \(g = f\) into the orthogonality condition:
\[
2g \cdot \frac{5}{4} + 2g \cdot \left(-\frac{3}{2}\right) = c + 2
\]
This simplifies to:
\[
\frac{5g}{2} - 3g = c + 2
\]
or
\[
-\frac{g}{2} = c + 2
\]
### Step 5: Substitute \(c\) from Previous Equations
From Equation 1, we have:
\[
c = -4 - 4g
\]
Substituting this into the orthogonality condition:
\[
-\frac{g}{2} = -4 - 4g + 2
\]
This simplifies to:
\[
-\frac{g}{2} = -2 - 4g
\]
Multiplying through by -2 to eliminate the fraction:
\[
g = 4 + 8g
\]
Rearranging gives:
\[
-7g = 4 \implies g = -\frac{4}{7}
\]
Since \(g = f\), we also have:
\[
f = -\frac{4}{7}
\]
### Step 6: Find \(c\)
Now substituting \(g\) back to find \(c\):
\[
c = -4 - 4\left(-\frac{4}{7}\right) = -4 + \frac{16}{7} = -\frac{28}{7} + \frac{16}{7} = -\frac{12}{7}
\]
### Step 7: Write the Final Equation
Now substituting \(g\), \(f\), and \(c\) back into the general equation:
\[
x^2 + y^2 + 2\left(-\frac{4}{7}\right)x + 2\left(-\frac{4}{7}\right)y - \frac{12}{7} = 0
\]
This simplifies to:
\[
x^2 + y^2 - \frac{8}{7}x - \frac{8}{7}y - \frac{12}{7} = 0
\]
Multiplying through by 7 to eliminate the fractions:
\[
7x^2 + 7y^2 - 8x - 8y - 12 = 0
\]
### Final Answer
The equation of the required circle is:
\[
7x^2 + 7y^2 - 8x - 8y - 12 = 0
\]
