Find the centre and radius of each of the
circles whose equations are given below.
`sqrt(1+m^(3) ) (x^(2)+y^(2))-2cx - 2mcy = 0`

Answer

To find the center and radius of the circle given by the equation:

\[ \sqrt{1 + m^3} (x^2 + y^2) - 2cx - 2mcy = 0 \]

we will follow these steps:

### Step 1: Rewrite the equation
First, we need to rewrite the equation in a standard form. We can do this by dividing the entire equation by \(\sqrt{1 + m^3}\):

\[ x^2 + y^2 - \frac{2c}{\sqrt{1 + m^3}} x - \frac{2mc}{\sqrt{1 + m^3}} y = 0 \]

### Step 2: Identify coefficients
Now, we can identify the coefficients in the standard form of the circle's equation, which is:

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

From our rewritten equation, we can see:
- \(g = -\frac{c}{\sqrt{1 + m^3}}\)
- \(f = -\frac{mc}{\sqrt{1 + m^3}}\)

### Step 3: Find the center
The center of the circle \((h, k)\) can be found using the formulas:
\[ h = -g = \frac{c}{\sqrt{1 + m^3}} \]
\[ k = -f = \frac{mc}{\sqrt{1 + m^3}} \]

Thus, the center of the circle is:
\[ \left(\frac{c}{\sqrt{1 + m^3}}, \frac{mc}{\sqrt{1 + m^3}}\right) \]

### Step 4: Find the radius
The radius \(r\) can be calculated using the formula:
\[ r = \sqrt{g^2 + f^2 - c} \]

Since there is no constant term in our equation, \(c = 0\). Therefore, we have:
\[ r = \sqrt{g^2 + f^2} \]

Substituting \(g\) and \(f\):
\[ r = \sqrt{\left(-\frac{c}{\sqrt{1 + m^3}}\right)^2 + \left(-\frac{mc}{\sqrt{1 + m^3}}\right)^2} \]
\[ = \sqrt{\frac{c^2}{1 + m^3} + \frac{m^2c^2}{1 + m^3}} \]
\[ = \sqrt{\frac{c^2(1 + m^2)}{1 + m^3}} \]
\[ = \frac{c\sqrt{1 + m^2}}{\sqrt{1 + m^3}} \]

### Final Result
Thus, the center and radius of the circle are:
- **Center:** \(\left(\frac{c}{\sqrt{1 + m^3}}, \frac{mc}{\sqrt{1 + m^3}}\right)\)
- **Radius:** \(\frac{c\sqrt{1 + m^2}}{\sqrt{1 + m^3}}\)

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