A circle of radius 2 units rolls inside the ring of the circle `x^(2)+y^(2)+8x-2y-19=0` then the locus of its centre is

Options

A. `x^(2)+y^(2)+8x+2y-47=0`
B. `x^(2)+y^(2)+8x-2y-1=0`
C. `x^(2)+y^(2)+8x-2y+1=0`
D. `x^(2)+y^(2)-8x+2y+1=0`

Answer

To find the locus of the center of a circle of radius 2 units rolling inside the ring of the circle given by the equation \(x^2 + y^2 + 8x - 2y - 19 = 0\), we will follow these steps:

### Step 1: Rewrite the equation of the larger circle
First, we need to rewrite the equation of the larger circle in standard form. The given equation is:

\[ x^2 + y^2 + 8x - 2y - 19 = 0 \]

We can rearrange it as follows:

\[ x^2 + 8x + y^2 - 2y = 19 \]

### Step 2: Complete the square
Next, we will complete the square for both the \(x\) and \(y\) terms.

For \(x\):
\[ x^2 + 8x = (x + 4)^2 - 16 \]

For \(y\):
\[ y^2 - 2y = (y - 1)^2 - 1 \]

Substituting these back into the equation gives:

\[ ((x + 4)^2 - 16) + ((y - 1)^2 - 1) = 19 \]

Simplifying this, we have:

\[ (x + 4)^2 + (y - 1)^2 - 17 = 19 \]

So,

\[ (x + 4)^2 + (y - 1)^2 = 36 \]

### Step 3: Identify the center and radius of the larger circle
From the equation \((x + 4)^2 + (y - 1)^2 = 36\), we can see that the center of the larger circle is at \((-4, 1)\) and the radius \(R\) is \(6\) (since \(6^2 = 36\)).

### Step 4: Determine the locus of the center of the smaller circle
The smaller circle has a radius of \(2\) units. When this circle rolls inside the larger circle, the distance from the center of the smaller circle to the center of the larger circle must be equal to the radius of the larger circle minus the radius of the smaller circle:

\[ PC = R - r = 6 - 2 = 4 \]

### Step 5: Set up the distance formula
Let the center of the smaller circle be \(P(x, y)\). The distance from \(P\) to the center of the larger circle \(C(-4, 1)\) is given by:

\[ PC = \sqrt{(x + 4)^2 + (y - 1)^2} \]

Setting this equal to \(4\):

\[ \sqrt{(x + 4)^2 + (y - 1)^2} = 4 \]

### Step 6: Square both sides
Squaring both sides gives:

\[ (x + 4)^2 + (y - 1)^2 = 16 \]

### Step 7: Expand and simplify
Expanding this equation:

\[ (x + 4)^2 + (y - 1)^2 = 16 \]

Expanding the left side:

\[ (x^2 + 8x + 16) + (y^2 - 2y + 1) = 16 \]

Combining like terms:

\[ x^2 + y^2 + 8x - 2y + 17 = 16 \]

Subtracting \(16\) from both sides:

\[ x^2 + y^2 + 8x - 2y + 1 = 0 \]

### Conclusion
Thus, the locus of the center of the smaller circle is given by the equation:

\[ x^2 + y^2 + 8x - 2y + 1 = 0 \]