The shortest distance between the circles `x ^(2) + y ^(2) =1 and (x-9)^(2) + (y-12)^(2) =4` is

Options

A. 9
B. 11
C. 12
D. 14

Answer

To find the shortest distance between the circles given by the equations \(x^2 + y^2 = 1\) and \((x - 9)^2 + (y - 12)^2 = 4\), we can follow these steps:

### Step 1: Identify the centers and radii of the circles
1. The first circle \(x^2 + y^2 = 1\) is centered at \(C_1(0, 0)\) with a radius \(r_1 = 1\).
2. The second circle \((x - 9)^2 + (y - 12)^2 = 4\) has its center at \(C_2(9, 12)\) with a radius \(r_2 = 2\).

### Step 2: Calculate the distance between the centers of the circles
The distance \(d\) between the centers \(C_1(0, 0)\) and \(C_2(9, 12)\) can be calculated using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ d = \sqrt{(9 - 0)^2 + (12 - 0)^2} = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15 \]

### Step 3: Calculate the shortest distance between the circles
The shortest distance \(D\) between the two circles can be found by subtracting the sum of the radii from the distance between the centers:
\[ D = d - (r_1 + r_2) \]
Substituting the values:
\[ D = 15 - (1 + 2) = 15 - 3 = 12 \]

### Conclusion
The shortest distance between the two circles is \(12\) units.

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