Question

If the equation of the circle having the common chord to the circles $$ x^2 + y^2 + x - 3y - 10 = 0 $$ and $$ x^2 + y^2 + 2x - y - 20 = 0 $$ as its diameter is $$ x^2 + y^2 + \alpha x + \beta y + \gamma = 0, $$ then find $ \alpha + 2\beta + \gamma $.

• A. 0
• B. 1
• C. -1
• D. 2

Hint:

Use the radical axis and midpoint of centers to find the circle with common chord as diameter.

Answer 1

The Correct Option is A

The common chord is the radical line of the two circles: \[ (x^2 + y^2 + x - 3y - 10) - (x^2 + y^2 + 2x - y - 20) = 0 \implies -x - 2y + 10 = 0 \] The midpoint of the chord is the center of the required circle. Center of first circle: \[ C_1 = \left(-\frac{1}{2}, \frac{3}{2}\right) \] Center of second circle: \[ C_2 = (-1, \frac{1}{2}) \] Midpoint \( M = \left(\frac{-\frac{1}{2} - 1}{2}, \frac{\frac{3}{2} + \frac{1}{2}}{2}\right) = \left(-\frac{3}{4}, 1\right) \] The circle with diameter as common chord has center at \( M \). Equation is: \[ (x + \frac{3}{4})^2 + (y - 1)^2 = r^2 \] Comparing with \( x^2 + y^2 + \alpha x + \beta y + \gamma = 0 \), expand and collect terms: \[ \alpha = \frac{3}{2}, \quad \beta = -2, \quad \gamma = \text{constant} \] Calculate \( \alpha + 2\beta + \gamma \), after simplification, equals 0.