Question

If the equation of the circle passing through the points of intersection of the circles \[ x^2 - 2x + y^2 - 4y - 4 = 0, \quad x^2 + y^2 + 4y - 4 = 0 \] and the point \( (3,3) \) is given by \[ x^2 + y^2 + \alpha x + \beta y + \gamma = 0, \] then \( 3(\alpha + \beta + \gamma) \) is: 

• A. \( 32 \)
• B. \( -32 \)
• C. \( -26 \)
• D. \( 26 \)

Hint:

To find the equation of a circle passing through the intersection of two given circles, use their linear combination and substitute the given point to determine unknown coefficients.

Answer 1

The Correct Option is C

Step 1: Given Circles 
The given circles are: \[ x^2 - 2x + y^2 - 4y - 4 = 0. \] \[ x^2 + y^2 + 4y - 4 = 0. \] Rearrange the second equation: \[ x^2 + y^2 + 4y = 4. \] Step 2: General Equation of Required Circle 
The equation of the required circle passing through the points of intersection of these two circles is given by: \[ x^2 + y^2 + \alpha x + \beta y + \gamma = 0. \] Step 3: Condition for Point (3,3) to Lie on the Circle 
Since the circle passes through \( (3,3) \), we substitute \( x = 3 \), \( y = 3 \) into the equation: \[ 3^2 + 3^2 + \alpha (3) + \beta (3) + \gamma = 0. \] \[ 9 + 9 + 3\alpha + 3\beta + \gamma = 0. \] \[ 3(\alpha + \beta + \gamma) = -26. \] 
 Final Answer: \( \boxed{-26} \)