The point `P(-1,0)` lies on the circle `x^(2)+y^(2)-4x+8y+k=0`. Find the radius of the circle Also determine the equation of the circle of equal radius which touches the given circle at P.

Answer

To solve the problem, we need to follow these steps:

### Step 1: Substitute the point P into the circle's equation to find k
Given the point \( P(-1, 0) \) lies on the circle defined by the equation:

\[ x^2 + y^2 - 4x + 8y + k = 0 \]

Substituting \( x = -1 \) and \( y = 0 \):

\[ (-1)^2 + (0)^2 - 4(-1) + 8(0) + k = 0 \]

This simplifies to:

\[ 1 + 0 + 4 + 0 + k = 0 \]

Thus,

\[ 5 + k = 0 \implies k = -5 \]

### Step 2: Write the equation of the circle
Now that we have \( k \), we can write the equation of the circle:

\[ x^2 + y^2 - 4x + 8y - 5 = 0 \]

### Step 3: Identify the center and radius of the circle
The general form of a circle's equation is:

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

From our equation, we can identify:

- \( 2g = -4 \) → \( g = -2 \)
- \( 2f = 8 \) → \( f = 4 \)
- \( c = -5 \)

The center \( (h, k) \) of the circle is given by \( (-g, -f) \):

\[ \text{Center} = (2, -4) \]

The radius \( r \) is calculated using the formula:

\[ r = \sqrt{g^2 + f^2 - c} \]

Substituting the values:

\[ r = \sqrt{(-2)^2 + (4)^2 - (-5)} = \sqrt{4 + 16 + 5} = \sqrt{25} = 5 \]

### Step 4: Find the equation of the new circle that touches the original circle at point P
Let the center of the new circle be \( (h, k) \). Since it touches the original circle at point \( P(-1, 0) \) and has the same radius \( r = 5 \), we can use the distance formula:

\[ \text{Distance} = \sqrt{(h + 1)^2 + (k - 0)^2} = 5 \]

Squaring both sides gives:

\[ (h + 1)^2 + k^2 = 25 \]

### Step 5: Rearranging the equation
Expanding the equation:

\[ h^2 + 2h + 1 + k^2 = 25 \]

Rearranging gives:

\[ h^2 + k^2 + 2h - 24 = 0 \]

### Step 6: Replace variables for the final equation
Let \( x = h \) and \( y = k \). The equation becomes:

\[ x^2 + y^2 + 2x - 24 = 0 \]

### Final Answer
The radius of the circle is \( 5 \), and the equation of the circle that touches the given circle at point \( P(-1, 0) \) is:

\[ x^2 + y^2 + 2x - 24 = 0 \]