Obtain the equation of the circle, centre (1, 0), which passes through the point P(3, 1 `(1)/(2)`). Find also the equation of the equal circle which touches the given circle externally at P.

Answer

To solve the problem, we need to find the equation of a circle with a given center and a point through which it passes. Then, we will find the equation of another circle that touches the first circle externally at the given point.

### Step 1: Identify the center and the point
The center of the circle is given as \( C(1, 0) \) and the point \( P(3, 1.5) \) through which the circle passes.

### Step 2: Calculate the radius
To find the radius \( r \) of the circle, we calculate the distance between the center \( C \) and the point \( P \) using the distance formula:
\[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates of \( C(1, 0) \) and \( P(3, 1.5) \):
\[ r = \sqrt{(3 - 1)^2 + (1.5 - 0)^2} = \sqrt{(2)^2 + (1.5)^2} = \sqrt{4 + 2.25} = \sqrt{6.25} = 2.5 \]

### Step 3: Write the equation of the circle
The standard form of the equation of a circle is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Where \( (h, k) \) is the center and \( r \) is the radius. Substituting \( h = 1 \), \( k = 0 \), and \( r = 2.5 \):
\[ (x - 1)^2 + (y - 0)^2 = (2.5)^2 \]
This simplifies to:
\[ (x - 1)^2 + y^2 = 6.25 \]

### Step 4: Expand the equation
Expanding the equation gives:
\[ (x - 1)^2 + y^2 = 6.25 \]
\[ x^2 - 2x + 1 + y^2 = 6.25 \]
\[ x^2 + y^2 - 2x + 1 - 6.25 = 0 \]
\[ x^2 + y^2 - 2x - 5.25 = 0 \]

### Step 5: Find the equation of the equal circle that touches the given circle externally
Let the center of the new circle be \( C' \) and its radius be \( r' \). Since the new circle touches the original circle externally at point \( P \), we have:
\[ CP + C'P = r + r' \]
Where \( CP \) is the radius of the original circle, which is \( 2.5 \), and \( C'P \) is the radius of the new circle \( r' \).

Since \( P \) is the midpoint between \( C \) and \( C' \):
\[ P = \left( \frac{1 + h}{2}, \frac{0 + k}{2} \right) \]
Substituting \( P(3, 1.5) \):
\[ 3 = \frac{1 + h}{2} \quad \text{and} \quad 1.5 = \frac{0 + k}{2} \]
From the first equation:
\[ 1 + h = 6 \implies h = 5 \]
From the second equation:
\[ 0 + k = 3 \implies k = 3 \]
Thus, the center of the new circle \( C'(5, 3) \).

### Step 6: Calculate the radius of the new circle
Since the new circle has the same radius as the original circle:
\[ r' = 2.5 \]

### Step 7: Write the equation of the new circle
Using the center \( C'(5, 3) \) and radius \( r' = 2.5 \):
\[ (x - 5)^2 + (y - 3)^2 = (2.5)^2 \]
This simplifies to:
\[ (x - 5)^2 + (y - 3)^2 = 6.25 \]

### Step 8: Expand the equation of the new circle
Expanding gives:
\[ (x - 5)^2 + (y - 3)^2 = 6.25 \]
\[ x^2 - 10x + 25 + y^2 - 6y + 9 = 6.25 \]
\[ x^2 + y^2 - 10x - 6y + 34 - 6.25 = 0 \]
\[ x^2 + y^2 - 10x - 6y + 27.75 = 0 \]

### Final Result
The equations of the circles are:
1. Original Circle: \( x^2 + y^2 - 2x - 5.25 = 0 \)
2. New Circle: \( x^2 + y^2 - 10x - 6y + 27.75 = 0 \)