If the two circles `x^(2)+y^(2)+2gx+c=0 and x^(2)+y^(2)-2fy-c=0` have equal radius then locus of (g,f) is

Options

A. a) `x^(2)+y^(2)=c^(2)`
B. b) `x^(2)-y^(2)=2c`
C. c) `x-y^(2)=c^(2)`
D. d) `x^(2)+y^(2)=2c^(2)`

Answer

To solve the problem, we need to find the locus of the points (g, f) such that the two given circles have equal radii.

### Step-by-Step Solution:

1. **Identify the equations of the circles**:
The equations of the two circles are:
- Circle 1: \( x^2 + y^2 + 2gx + c = 0 \)
- Circle 2: \( x^2 + y^2 - 2fy - c = 0 \)

2. **Recall the formula for the radius of a circle**:
The radius \( R \) of a circle given by the equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \) is given by:
\[ R = \sqrt{g^2 + f^2 - c} \]

3. **Calculate the radius of Circle 1**:
For Circle 1, the radius \( R_1 \) is:
\[ R_1 = \sqrt{g^2 - c} \]
(since there is no \( y \) term, \( f = 0 \)).

4. **Calculate the radius of Circle 2**:
For Circle 2, the radius \( R_2 \) is:
\[ R_2 = \sqrt{f^2 + c} \]
(since there is no \( x \) term, \( g = 0 \)).

5. **Set the radii equal**:
According to the problem, the radii of both circles are equal:
\[ R_1 = R_2 \]
This gives us:
\[ \sqrt{g^2 - c} = \sqrt{f^2 + c} \]

6. **Square both sides**:
Squaring both sides to eliminate the square roots:
\[ g^2 - c = f^2 + c \]

7. **Rearrange the equation**:
Rearranging gives:
\[ g^2 - f^2 = 2c \]

8. **Replace variables**:
Let \( g = x \) and \( f = y \) to express the locus:
\[ x^2 - y^2 = 2c \]

### Final Result:
The locus of the points \( (g, f) \) is given by the equation:
\[ x^2 - y^2 = 2c \]