The external centre of similitude of the two circles `x^(2)+y^(2)-2x-6y+9=0, x^(2)+y^(2)=4` is

Options

A. (1,3)
B. (2,6)
C. (3,9)
D. (1,2)

Answer

To find the external center of similitude of the two circles given by the equations \(x^2 + y^2 - 2x - 6y + 9 = 0\) and \(x^2 + y^2 = 4\), we will follow these steps:

### Step 1: Rewrite the equations of the circles in standard form.

1. For the first circle:
\[ x^2 + y^2 - 2x - 6y + 9 = 0 \]
We can complete the square for \(x\) and \(y\):
\[ (x^2 - 2x) + (y^2 - 6y) + 9 = 0 \]
Completing the square:
\[ (x - 1)^2 - 1 + (y - 3)^2 - 9 + 9 = 0 \]
This simplifies to:
\[ (x - 1)^2 + (y - 3)^2 = 1 \]
So, the center \(C_1\) is \((1, 3)\) and the radius \(r_1 = 1\).

2. For the second circle:
\[ x^2 + y^2 = 4 \]
This is already in standard form:
\[ (x - 0)^2 + (y - 0)^2 = 2^2 \]
So, the center \(C_2\) is \((0, 0)\) and the radius \(r_2 = 2\).

### Step 2: Identify the centers and radii.

From the above steps:
- Center of the first circle \(C_1 = (1, 3)\) and radius \(r_1 = 1\).
- Center of the second circle \(C_2 = (0, 0)\) and radius \(r_2 = 2\).

### Step 3: Use the formula for the external center of similitude.

The formula for the external center of similitude is given by:
\[ x = \frac{x_2 r_1 - x_1 r_2}{r_1 - r_2} \]
\[ y = \frac{y_2 r_1 - y_1 r_2}{r_1 - r_2} \]

Substituting the values:
- \(x_1 = 1\), \(y_1 = 3\), \(x_2 = 0\), \(y_2 = 0\), \(r_1 = 1\), \(r_2 = 2\).

### Step 4: Calculate the x-coordinate.

\[ x = \frac{0 \cdot 1 - 1 \cdot 2}{1 - 2} = \frac{-2}{-1} = 2 \]

### Step 5: Calculate the y-coordinate.

\[ y = \frac{0 \cdot 1 - 3 \cdot 2}{1 - 2} = \frac{-6}{-1} = 6 \]

### Step 6: Conclusion.

The external center of similitude is \((2, 6)\).