A point Plies inside the circles `x^2+y^2-4=0 and x^2+y^2-8x+7=0`. The poirt P starts moving such that it is always inside the circles, its path enclosus greatest possible area and it is at a fixeddistance from an arbitrarily chosen point in its region. The locus of P is.

Options

A. (a)`4x^(2)+4y^(2)-12x-1=0`
B. (b)`4x^(2)+4y^(2)+12x+1=0`
C. (c)`4x^(2)+4y^(2)-3x-2=0`
D. (d)`4x^(2)+4y^(2)-3x+2=0`

Answer

To solve the problem step-by-step, we need to find the locus of the point P that lies inside the two given circles and moves in such a way that it encloses the greatest possible area while maintaining a fixed distance from a chosen point within its region.

### Step 1: Write the equations of the circles
The first circle is given by:
\[ x^2 + y^2 - 4 = 0 \implies x^2 + y^2 = 4 \]
This represents a circle centered at (0, 0) with a radius of 2.

The second circle is given by:
\[ x^2 + y^2 - 8x + 7 = 0 \]
We can rewrite this equation by completing the square:
\[ x^2 - 8x + y^2 + 7 = 0 \implies (x - 4)^2 + y^2 = 16 \]
This represents a circle centered at (4, 0) with a radius of 4.

### Step 2: Identify the centers and radii
From the above equations:
- Circle 1: Center \(C_1(0, 0)\), Radius \(r_1 = 2\)
- Circle 2: Center \(C_2(4, 0)\), Radius \(r_2 = 4\)

### Step 3: Determine the area of interest
The point P must lie within both circles. The area of interest is the region where both circles overlap. The maximum area that P can enclose while maintaining a fixed distance from a point within this region will be a smaller circle.

### Step 4: Find the center of the smaller circle
To maximize the area while remaining within both circles, the center of the smaller circle should be positioned at the midpoint between the centers of the two circles. The midpoint \(M\) between \(C_1\) and \(C_2\) is:
\[ M = \left(\frac{0 + 4}{2}, \frac{0 + 0}{2}\right) = (2, 0) \]

### Step 5: Determine the radius of the smaller circle
The radius of the smaller circle can be determined by the distance from the center \(M(2, 0)\) to the boundary of the smaller circle, which must be at a fixed distance from the chosen point. The distance from \(M\) to the edge of the first circle (radius 2) is:
\[ \text{Distance from } M \text{ to } C_1 = 2 - 2 = 0 \]
And the distance from \(M\) to the edge of the second circle (radius 4) is:
\[ \text{Distance from } M \text{ to } C_2 = 4 - 2 = 2 \]
Thus, the radius of the smaller circle is half of the distance between the two circles, which is:
\[ \text{Radius of smaller circle} = \frac{1}{2} \]

### Step 6: Write the equation of the smaller circle
The equation of the smaller circle centered at \(M(2, 0)\) with radius \(\frac{1}{2}\) is:
\[ (x - 2)^2 + y^2 = \left(\frac{1}{2}\right)^2 \]
This simplifies to:
\[ (x - 2)^2 + y^2 = \frac{1}{4} \]

### Final Answer
The locus of point P is given by:
\[ (x - 2)^2 + y^2 = \frac{1}{4} \]