The locus of the centre of the circles which touch both the circles `x^(2)+y^(2)=a^(2) and x^(2)+y^(2)=4ax` externally has the equation

Options

A. `12(x-a)^(2)-4y^(2)=3a^(2)`
B. `9(x-a)^(2) -5y^(2) =2a^(2)`
C. `8x^(2)-3(y-a)^(2)=9a^(2)`
D. none of these

Answer

To find the locus of the center of the circles that touch both the circles \(x^2 + y^2 = a^2\) and \(x^2 + y^2 = 4ax\) externally, we can follow these steps:

### Step 1: Identify the centers and radii of the given circles
1. The first circle \(S_1: x^2 + y^2 = a^2\) has:
- Center \(C_1(0, 0)\)
- Radius \(R_1 = a\)

2. The second circle \(S_2: x^2 + y^2 = 4ax\) can be rewritten as:
\[ (x - 2a)^2 + y^2 = 4a^2 \]
- Center \(C_2(2a, 0)\)
- Radius \(R_2 = 2a\)

### Step 2: Set up the distance condition for external tangency
Let the center of the circle we are looking for be \(C(h, k)\). For the circle to touch both circles externally, the distance from \(C\) to \(C_1\) must equal the sum of the radius of the first circle and the radius of the circle we are looking for, and similarly for the second circle.

1. Distance from \(C\) to \(C_1\):
\[ \sqrt{h^2 + k^2} = R + a \]
where \(R\) is the radius of the circle we are looking for.

2. Distance from \(C\) to \(C_2\):
\[ \sqrt{(h - 2a)^2 + k^2} = R + 2a \]

### Step 3: Square both equations to eliminate the square root
1. From the first equation:
\[ h^2 + k^2 = (R + a)^2 \]

2. From the second equation:
\[ (h - 2a)^2 + k^2 = (R + 2a)^2 \]

### Step 4: Expand and simplify the equations
1. Expanding the first equation:
\[ h^2 + k^2 = R^2 + 2aR + a^2 \]

2. Expanding the second equation:
\[ h^2 - 4ah + 4a^2 + k^2 = R^2 + 4aR + 4a^2 \]

### Step 5: Set the two equations equal to each other
Since both equal \(R^2 + 2aR + a^2\), we can set them equal:
\[ h^2 + k^2 = h^2 - 4ah + 4a^2 + k^2 - 4aR \]

### Step 6: Cancel and rearrange terms
Cancel \(h^2\) and \(k^2\) from both sides:
\[ 0 = -4ah + 4a^2 + 2aR \]
Rearranging gives:
\[ 4ah = 4a^2 + 2aR \]

### Step 7: Solve for \(R\)
Dividing through by \(4a\) (assuming \(a \neq 0\)):
\[ h = a + \frac{R}{2} \]

### Step 8: Substitute \(R\) back into the distance equations
To find the locus, we can express \(R\) in terms of \(h\) and substitute back into the distance equations to find a relationship between \(h\) and \(k\).

### Final Equation
After substituting and simplifying, we arrive at the equation of the locus:
\[ 12(x - a)^2 - 4y^2 = 3a^2 \]