Consider the equation of circle `x^(2)+y^(2)-2x-2lambday -8=0` where `lambda` is variable then answer the following:
The given equation represents a family of circles passing through two fixed points. Find the coordinates of the fixed points.

Answer

To solve the problem, we start with the given equation of the circle:

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

### Step 1: Rearranging the Equation
We can rearrange the equation to isolate the terms involving \(x\) and \(y\):

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

### Step 2: Completing the Square
Next, we complete the square for the \(x\) and \(y\) terms.

For \(x\):
\[ x^2 - 2x = (x - 1)^2 - 1 \]

For \(y\):
\[ y^2 - 2\lambda y = (y - \lambda)^2 - \lambda^2 \]

Substituting these back into the equation gives:

\[ (x - 1)^2 - 1 + (y - \lambda)^2 - \lambda^2 - 8 = 0 \]

### Step 3: Simplifying the Equation
Now, we simplify the equation:

\[ (x - 1)^2 + (y - \lambda)^2 - 1 - \lambda^2 - 8 = 0 \]

This simplifies to:

\[ (x - 1)^2 + (y - \lambda)^2 = \lambda^2 + 9 \]

### Step 4: Identifying the Fixed Points
The equation \((x - 1)^2 + (y - \lambda)^2 = \lambda^2 + 9\) represents a family of circles centered at \((1, \lambda)\) with radius \(\sqrt{\lambda^2 + 9}\).

To find the fixed points through which all these circles pass, we need to analyze the situation when \(\lambda\) takes specific values.

### Step 5: Finding Fixed Points
To find the fixed points, we can set \(\lambda = k\) for some constant \(k\) and analyze the resulting equations.

Setting \(\lambda = 0\):

\[ (x - 1)^2 + y^2 = 9 \]

This circle has a center at \((1, 0)\) and a radius of \(3\).

Setting \(\lambda = 3\):

\[ (x - 1)^2 + (y - 3)^2 = 18 \]

This circle has a center at \((1, 3)\) and a radius of \(3\sqrt{2}\).

The two fixed points that all circles pass through can be found by setting \(y\) to specific values (like \(y = 0\) and \(y = 3\)).

### Final Fixed Points
The fixed points are:

1. \((1, 0)\)
2. \((1, 3)\)

Thus, the coordinates of the fixed points through which the family of circles passes are \((1, 0)\) and \((1, 3)\).