Show that the circles
`x^(2)+y^(2)-6x+4y+4=0and x^(2)+y^(2)+x+4y+1=0` cut orthogonally.

Answer

To show that the circles given by the equations \( x^2 + y^2 - 6x + 4y + 4 = 0 \) and \( x^2 + y^2 + x + 4y + 1 = 0 \) cut orthogonally, we will follow these steps:

### Step 1: Identify the coefficients of the circles
The general equation of a circle is given by:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From the first circle \( S_1: x^2 + y^2 - 6x + 4y + 4 = 0 \):
- \( g_1 = -3 \)
- \( f_1 = 2 \)
- \( c_1 = 4 \)

From the second circle \( S_2: x^2 + y^2 + x + 4y + 1 = 0 \):
- \( g_2 = \frac{1}{2} \)
- \( f_2 = 2 \)
- \( c_2 = 1 \)

### Step 2: Use the orthogonality condition
The circles cut orthogonally if the following condition holds:
\[ 2g_1g_2 + 2f_1f_2 = c_1 + c_2 \]

### Step 3: Calculate the left-hand side (LHS)
Substituting the values into the LHS:
\[ LHS = 2g_1g_2 + 2f_1f_2 = 2(-3)\left(\frac{1}{2}\right) + 2(2)(2) \]
Calculating each term:
\[ = 2(-3)(\frac{1}{2}) + 2(2)(2) = -3 + 8 = 5 \]

### Step 4: Calculate the right-hand side (RHS)
Now we calculate the RHS:
\[ RHS = c_1 + c_2 = 4 + 1 = 5 \]

### Step 5: Compare LHS and RHS
Since \( LHS = 5 \) and \( RHS = 5 \), we have:
\[ LHS = RHS \]

### Conclusion
Since the LHS equals the RHS, the circles cut orthogonally.

Thus, we have shown that the circles \( x^2 + y^2 - 6x + 4y + 4 = 0 \) and \( x^2 + y^2 + x + 4y + 1 = 0 \) cut orthogonally.

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