If the circle `(x-2) ^(2) + (y -3) ^(2)=a ^(2)` lies entirely in the circle `x ^(2) + y ^(2) =b ^(2),` then

Options

A. `a=b`
B. `a -b gt sqrt13`
C. `b -a gt sqrt13`
D. `b-a = sqrt13`

Answer

To solve the problem, we need to determine the conditions under which the circle \((x-2)^2 + (y-3)^2 = a^2\) lies entirely within the circle \(x^2 + y^2 = b^2\).

### Step 1: Identify the centers and radii of the circles

The first circle has the equation:
\[ (x-2)^2 + (y-3)^2 = a^2 \]
- Center: \((2, 3)\)
- Radius: \(a\)

The second circle has the equation:
\[ x^2 + y^2 = b^2 \]
- Center: \((0, 0)\)
- Radius: \(b\)

### Step 2: Calculate the distance between the centers of the circles

The distance \(d\) between the centers \((2, 3)\) and \((0, 0)\) is given by the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ d = \sqrt{(0 - 2)^2 + (0 - 3)^2} = \sqrt{(-2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} \]

### Step 3: Establish the condition for one circle to lie entirely within the other

For the first circle to lie entirely within the second circle, the following condition must be satisfied:
\[ \text{Radius of second circle} - \text{Radius of first circle} > \text{Distance between centers} \]
This can be expressed mathematically as:
\[ b - a > d \]
Substituting the distance we calculated:
\[ b - a > \sqrt{13} \]

### Conclusion

Thus, the condition for the circle \((x-2)^2 + (y-3)^2 = a^2\) to lie entirely within the circle \(x^2 + y^2 = b^2\) is:
\[ b - a > \sqrt{13} \]