The radius of a circle is 5 units and it touches the circle `x^(2) + y^(2) - 2 x - 4 y - 2 0 = 0` externally at the point (5,5). Find the equation of the circle.

Answer

To find the equation of the circle that touches the given circle externally at the point (5, 5) and has a radius of 5 units, we can follow these steps:

### Step 1: Write the equation of the given circle in standard form
The given equation is:
\[ x^2 + y^2 - 2x - 4y - 20 = 0 \]

To convert it into standard form, we will complete the square for both \(x\) and \(y\).

1. Rearranging the equation:
\[ x^2 - 2x + y^2 - 4y = 20 \]

2. Completing the square:
- For \(x^2 - 2x\):
\[ x^2 - 2x = (x - 1)^2 - 1 \]
- For \(y^2 - 4y\):
\[ y^2 - 4y = (y - 2)^2 - 4 \]

3. Substitute back into the equation:
\[ (x - 1)^2 - 1 + (y - 2)^2 - 4 = 20 \]
\[ (x - 1)^2 + (y - 2)^2 = 25 \]

So, the center of the given circle \(C_1\) is \((1, 2)\) and its radius \(r_1\) is \(5\).

### Step 2: Determine the center of the new circle
Let the center of the new circle be \((H, K)\). Since the new circle touches the given circle externally at the point (5, 5), we can use the midpoint theorem.

The distance between the centers of the two circles is equal to the sum of their radii:
\[ \text{Distance} = r_1 + r_2 = 5 + 5 = 10 \]

### Step 3: Use the midpoint theorem
The midpoint of the line segment joining the centers \(C_1(1, 2)\) and \(C_2(H, K)\) is the point of tangency (5, 5). Therefore, we can set up the equations:
1. For the x-coordinates:
\[ \frac{1 + H}{2} = 5 \]
\[ 1 + H = 10 \]
\[ H = 9 \]

2. For the y-coordinates:
\[ \frac{2 + K}{2} = 5 \]
\[ 2 + K = 10 \]
\[ K = 8 \]

Thus, the center of the new circle \(C_2\) is \((9, 8)\).

### Step 4: Write the equation of the new circle
The standard form of the equation of a circle is given by:
\[ (x - H)^2 + (y - K)^2 = r^2 \]

Substituting \(H = 9\), \(K = 8\), and \(r = 5\):
\[ (x - 9)^2 + (y - 8)^2 = 5^2 \]
\[ (x - 9)^2 + (y - 8)^2 = 25 \]

### Final Answer
The equation of the circle is:
\[ (x - 9)^2 + (y - 8)^2 = 25 \]

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