Find the equation of the circle which cuts each of the circles `x^2+y^2=4`, `x^2 +y^2-6x-8y.+ 10=0` & `x^2 + y^2+2x-4y-2 = 0` at the extremities of a diameter

Options

A. c=-4
B. g+f=c=-1
C. `g^(2)+f^(2)-c=17`
D. gf=6

Answer

To find the equation of the circle that intersects each of the given circles at the extremities of a diameter, we will follow these steps:

### Step 1: Write the equations of the given circles
The equations of the circles are:
1. \( S_1: x^2 + y^2 = 4 \)
2. \( S_2: x^2 + y^2 - 6x - 8y + 10 = 0 \)
3. \( S_3: x^2 + y^2 + 2x - 4y - 2 = 0 \)

### Step 2: Rewrite the equations in standard form
- For \( S_1 \), it is already in standard form.
- For \( S_2 \):
\[ S_2: x^2 + y^2 - 6x - 8y + 10 = 0 \implies (x - 3)^2 + (y - 4)^2 = 9 \]
This circle has center \( (3, 4) \) and radius \( 3 \).

- For \( S_3 \):
\[ S_3: x^2 + y^2 + 2x - 4y - 2 = 0 \implies (x + 1)^2 + (y - 2)^2 = 5 \]
This circle has center \( (-1, 2) \) and radius \( \sqrt{5} \).

### Step 3: Set up the general equation of the circle
Let the equation of the required circle be:
\[ S: x^2 + y^2 + 2gx + 2fy + c = 0 \]

### Step 4: Use the condition for the common chord
For the circle \( S \) to intersect the circle \( S_1 \) at the extremities of a diameter, we have:
\[ S - S_1 = 0 \implies (2g)x + (2f)y + c + 4 = 0 \]
This line must pass through the center of \( S_1 \), which is \( (0, 0) \):
\[ c + 4 = 0 \implies c = -4 \]

### Step 5: Apply the same condition for \( S_2 \)
For \( S \) to intersect \( S_2 \):
\[ S - S_2 = 0 \implies (2g)x + (2f)y + c - 6x - 8y + 10 = 0 \]
Substituting \( c = -4 \):
\[ (2g - 6)x + (2f - 8)y + 6 = 0 \]
This line must pass through the center of \( S_2 \) at \( (3, 4) \):
\[ (2g - 6)(3) + (2f - 8)(4) + 6 = 0 \]
Expanding this gives:
\[ 6g - 18 + 8f - 32 + 6 = 0 \implies 6g + 8f - 44 = 0 \implies 3g + 4f = 22 \quad \text{(Equation 1)} \]

### Step 6: Apply the same condition for \( S_3 \)
For \( S \) to intersect \( S_3 \):
\[ S - S_3 = 0 \implies (2g)x + (2f)y + c - 2x + 4y + 2 = 0 \]
Substituting \( c = -4 \):
\[ (2g - 2)x + (2f + 4)y - 2 = 0 \]
This line must pass through the center of \( S_3 \) at \( (-1, 2) \):
\[ (2g - 2)(-1) + (2f + 4)(2) - 2 = 0 \]
Expanding this gives:
\[ -2g + 2 + 4f + 8 - 2 = 0 \implies -2g + 4f + 8 = 0 \implies 2g - 4f = 8 \quad \text{(Equation 2)} \]

### Step 7: Solve the system of equations
We have:
1. \( 3g + 4f = 22 \)
2. \( 2g - 4f = 8 \)

Adding these two equations:
\[ (3g + 4f) + (2g - 4f) = 22 + 8 \implies 5g = 30 \implies g = 6 \]

Substituting \( g = 6 \) into Equation 1:
\[ 3(6) + 4f = 22 \implies 18 + 4f = 22 \implies 4f = 4 \implies f = 1 \]

### Step 8: Write the final equation of the circle
Now substituting \( g = 6 \), \( f = 1 \), and \( c = -4 \) into the general circle equation:
\[ x^2 + y^2 + 12x + 2y - 4 = 0 \]

### Final Answer:
The equation of the circle is:
\[ x^2 + y^2 + 12x + 2y - 4 = 0 \]