The circles `ax^(2)+ay^(2)+2g_(1)x+2f_(1)y+c_(1)=0" and "bx^(2)+by^(2)+2g_(2)x+2f_(2)y+c_(2)=0`
`(ane0and bne0)` cut orthogonally, if

Options

A. an ellipse
B. the radical axis of the given circles
C. a conic
D. another circle

Answer

To find the condition under which the circles given by the equations

1. \( ax^2 + ay^2 + 2g_1x + 2f_1y + c_1 = 0 \)
2. \( bx^2 + by^2 + 2g_2x + 2f_2y + c_2 = 0 \)

cut orthogonally, we follow these steps:

### Step 1: Rewrite the Circle Equations
We start by rewriting both circle equations in standard form.

For the first circle, divide the entire equation by \( a \):
\[ x^2 + y^2 + \frac{2g_1}{a}x + \frac{2f_1}{a}y + \frac{c_1}{a} = 0 \]
Let \( g_1' = \frac{g_1}{a} \), \( f_1' = \frac{f_1}{a} \), and \( c_1' = \frac{c_1}{a} \). Thus, we have:
\[ x^2 + y^2 + 2g_1'x + 2f_1'y + c_1' = 0 \]

For the second circle, divide the entire equation by \( b \):
\[ x^2 + y^2 + \frac{2g_2}{b}x + \frac{2f_2}{b}y + \frac{c_2}{b} = 0 \]
Let \( g_2' = \frac{g_2}{b} \), \( f_2' = \frac{f_2}{b} \), and \( c_2' = \frac{c_2}{b} \). Thus, we have:
\[ x^2 + y^2 + 2g_2'x + 2f_2'y + c_2' = 0 \]

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

### Step 3: Substitute the Values
Substituting the values of \( g_1, f_1, c_1 \) from the first circle and \( g_2, f_2, c_2 \) from the second circle, we have:
\[ 2 \left(\frac{g_1}{a}\right)\left(\frac{g_2}{b}\right) + 2 \left(\frac{f_1}{a}\right)\left(\frac{f_2}{b}\right) = \frac{c_1}{a} + \frac{c_2}{b} \]

### Step 4: Simplifying the Equation
Multiplying through by \( ab \) to eliminate the denominators gives:
\[ 2g_1g_2 + 2f_1f_2 = b c_1 + a c_2 \]

### Conclusion
Thus, the required condition for the circles to cut orthogonally is:
\[ 2g_1g_2 + 2f_1f_2 = b c_1 + a c_2 \]