Question

The equation of the circle is 3x²+3y²+6x-4y-1=0. Then its radius is

• A. \(\frac{1}{3}\)
• B. \(\frac{4}{3}\)
• C. \(\frac{2}{3}\)
• D. \(\frac{16}{3}\)
• E. \(\frac{8}{3}\)

Answer 1

The Correct Option is B

Step 1: Convert the given equation into standard form 
The given equation of the circle is: \[ 3x^2 + 3y^2 + 6x - 4y - 1 = 0 \] Divide the entire equation by 3 to simplify: \[ x^2 + y^2 + 2x - \frac{4}{3}y - \frac{1}{3} = 0 \]

Step 2: Complete the square  
Group the terms involving \( x \) and \( y \): \[ (x^2 + 2x) + (y^2 - \frac{4}{3}y) = \frac{1}{3} \] 
Completing the square for \( x^2 + 2x \): \[ x^2 + 2x = (x + 1)^2 - 1 \] 
Completing the square for \( y^2 - \frac{4}{3}y \): \[ y^2 - \frac{4}{3}y = \left(y - \frac{2}{3}\right)^2 - \frac{4}{9} \]

Step 3: Write the equation in standard form 
Substituting these values: \[ (x+1)^2 - 1 + (y - \frac{2}{3})^2 - \frac{4}{9} = \frac{1}{3} \] 
Simplifying: \[ (x+1)^2 + (y - \frac{2}{3})^2 = \frac{1}{3} + 1 + \frac{4}{9} \] \[ = \frac{3}{9} + \frac{9}{9} + \frac{4}{9} = \frac{16}{9} \]

Step 4: Find the radius 
The standard form of a circle is: \[ (x - h)^2 + (y - k)^2 = r^2 \] Comparing, we get: \[ r^2 = \frac{16}{9} \] \[ r = \frac{4}{3} \]

Final Answer: The radius of the circle is \(\frac{4}{3}\).