Find the coordinates of the centres and the radii of the circles whose equations are
`sqrt( 1 + m^(2)) ( x^(2) + y^(2)) - 2cx - 2mcy =0`

Answer

To find the coordinates of the centers and the radii of the circles given by the equation:

\[ \sqrt{1 + m^2} (x^2 + y^2) - 2cx - 2mcy = 0 \]

we will follow these steps:

### Step 1: Rewrite the Equation
We start with the given equation:

\[ \sqrt{1 + m^2} (x^2 + y^2) - 2cx - 2mcy = 0 \]

To simplify, we divide the entire equation by \(\sqrt{1 + m^2}\):

\[ x^2 + y^2 - \frac{2c}{\sqrt{1 + m^2}} x - \frac{2mc}{\sqrt{1 + m^2}} y = 0 \]

### Step 2: Compare with Standard Circle Equation
The standard form of a circle's equation is:

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

From our equation, we can identify:

- \(2g = -\frac{2c}{\sqrt{1 + m^2}}\)
- \(2f = -\frac{2mc}{\sqrt{1 + m^2}}\)
- \(c = 0\)

### Step 3: Solve for g and f
From the equations for \(g\) and \(f\):

\[ g = -\frac{c}{\sqrt{1 + m^2}} \]
\[ f = -\frac{mc}{\sqrt{1 + m^2}} \]

### Step 4: Find the Center Coordinates
The center of the circle is given by the coordinates \((-g, -f)\):

\[ \text{Center} = \left(\frac{c}{\sqrt{1 + m^2}}, \frac{mc}{\sqrt{1 + m^2}}\right) \]

### Step 5: Find the Radius
The radius \(r\) of the circle can be calculated using the formula:

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

Substituting the values of \(g\) and \(f\):

\[ r = \sqrt{\left(-\frac{c}{\sqrt{1 + m^2}}\right)^2 + \left(-\frac{mc}{\sqrt{1 + m^2}}\right)^2 - 0} \]

Calculating \(g^2 + f^2\):

\[ g^2 + f^2 = \frac{c^2}{1 + m^2} + \frac{m^2c^2}{1 + m^2} = \frac{c^2(1 + m^2)}{1 + m^2} = c^2 \]

Thus, the radius becomes:

\[ r = \sqrt{c^2} = |c| \]

Since the radius cannot be negative, we take:

\[ r = c \]

### Final Result
The coordinates of the center and the radius of the circle are:

- **Center**: \(\left(\frac{c}{\sqrt{1 + m^2}}, \frac{mc}{\sqrt{1 + m^2}}\right)\)
- **Radius**: \(c\)