The radius of the circle
`3x ^(2) + by ^(2) + 4 bx - 6by + b ^(2) =0` is

Options

A. 1
B. 3
C. `sqrt10`
D. `sqrt11`

Answer

To find the radius of the circle given by the equation \(3x^2 + by^2 + 4bx - 6by + b^2 = 0\), we will first rewrite the equation in the standard form of a circle and then apply the formula for the radius.

### Step-by-Step Solution:

1. **Identify the General Form of the Circle:**
The general equation of a circle is given by:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
where the radius \(r\) can be calculated using the formula:
\[ r = \sqrt{g^2 + f^2 - c} \]

2. **Rewrite the Given Equation:**
The given equation is:
\[ 3x^2 + by^2 + 4bx - 6by + b^2 = 0 \]
To convert this into the standard form, we can divide the entire equation by \(b\) (assuming \(b \neq 0\)):
\[ \frac{3}{b}x^2 + y^2 + \frac{4}{b}x - \frac{6}{b}y + 1 = 0 \]

3. **Identify Coefficients:**
From the rewritten equation, we can identify:
- \(A = \frac{3}{b}\)
- \(B = 1\)
- \(C = \frac{4}{b}\)
- \(D = -\frac{6}{b}\)
- \(E = 1\)

Now, we can express the equation in the standard form:
\[ \frac{3}{b}x^2 + y^2 + \frac{4}{b}x - \frac{6}{b}y + 1 = 0 \]
This can be rearranged to:
\[ x^2 + \frac{b}{3}y^2 + \frac{4}{3}x - 2y + \frac{b}{3} = 0 \]

4. **Extract Values of \(g\), \(f\), and \(c\):**
Comparing with the standard form:
- \(2g = \frac{4}{3} \implies g = \frac{2}{3}\)
- \(2f = -2 \implies f = -1\)
- \(c = \frac{b}{3}\)

5. **Calculate the Radius:**
Now substituting \(g\), \(f\), and \(c\) into the radius formula:
\[ r = \sqrt{g^2 + f^2 - c} \]
\[ r = \sqrt{\left(\frac{2}{3}\right)^2 + (-1)^2 - \frac{b}{3}} \]
\[ r = \sqrt{\frac{4}{9} + 1 - \frac{b}{3}} \]
\[ r = \sqrt{\frac{4}{9} + \frac{9}{9} - \frac{b}{3}} = \sqrt{\frac{13}{9} - \frac{b}{3}} \]

6. **Final Expression for Radius:**
Thus, the radius of the circle is:
\[ r = \sqrt{\frac{13 - 3b}{9}} = \frac{\sqrt{13 - 3b}}{3} \]