Find the equation of the circle passing through the origin, having its centre on the line x+y=4 and intersecting the circle `x^2+y^2-4x+2y+4=0` orthogonally.

Answer

To find the equation of the circle passing through the origin, having its center on the line \( x + y = 4 \), and intersecting the circle \( x^2 + y^2 - 4x + 2y + 4 = 0 \) orthogonally, we can follow these steps:

### Step 1: General Equation of the Circle
The general equation of a circle can be expressed as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
where \( ( -g, -f ) \) is the center of the circle.

### Step 2: Circle Passing Through the Origin
Since the circle passes through the origin \( (0, 0) \), we substitute \( x = 0 \) and \( y = 0 \) into the equation:
\[ 0^2 + 0^2 + 2g(0) + 2f(0) + c = 0 \]
This simplifies to:
\[ c = 0 \]
Thus, the equation of the circle simplifies to:
\[ x^2 + y^2 + 2gx + 2fy = 0 \]

### Step 3: Center on the Line \( x + y = 4 \)
The center of the circle \( (-g, -f) \) lies on the line \( x + y = 4 \). Therefore, we have:
\[ -g - f = 4 \quad \text{(Equation 1)} \]

### Step 4: Orthogonality Condition
The given circle can be rewritten in standard form:
\[ (x - 2)^2 + (y + 1)^2 = 1 \]
From this, we can identify:
- \( g' = 2 \)
- \( f' = -1 \)
- \( c' = 1 \)

The condition for orthogonality of two circles is given by:
\[ 2gg' + 2ff' = c + c' \]
Substituting the values we have:
\[ 2g(2) + 2f(-1) = 0 + 4 \]
This simplifies to:
\[ 4g - 2f = 4 \quad \text{(Equation 2)} \]

### Step 5: Solving the System of Equations
Now we have two equations:
1. \( -g - f = 4 \) (Equation 1)
2. \( 4g - 2f = 4 \) (Equation 2)

From Equation 1, we can express \( f \) in terms of \( g \):
\[ f = -g - 4 \]

Substituting this into Equation 2:
\[ 4g - 2(-g - 4) = 4 \]
This simplifies to:
\[ 4g + 2g + 8 = 4 \]
\[ 6g + 8 = 4 \]
\[ 6g = 4 - 8 \]
\[ 6g = -4 \quad \Rightarrow \quad g = -\frac{2}{3} \]

Now substituting \( g \) back into Equation 1 to find \( f \):
\[ -f - \left(-\frac{2}{3}\right) = 4 \]
\[ -f + \frac{2}{3} = 4 \]
\[ -f = 4 - \frac{2}{3} \]
Converting 4 to a fraction:
\[ -f = \frac{12}{3} - \frac{2}{3} = \frac{10}{3} \]
Thus:
\[ f = -\frac{10}{3} \]

### Step 6: Final Equation of the Circle
Now we have \( g = -\frac{2}{3} \), \( f = -\frac{10}{3} \), and \( c = 0 \). The equation of the circle becomes:
\[ x^2 + y^2 + 2\left(-\frac{2}{3}\right)x + 2\left(-\frac{10}{3}\right)y = 0 \]
This simplifies to:
\[ x^2 + y^2 - \frac{4}{3}x - \frac{20}{3}y = 0 \]
Multiplying through by 3 to eliminate the fractions:
\[ 3x^2 + 3y^2 - 4x - 20y = 0 \]

### Final Answer
The equation of the required circle is:
\[ 3x^2 + 3y^2 - 4x - 20y = 0 \]