Question

Find the values of \( \alpha \) for which the circle \[ x^2 + y^2 + \alpha x - 8y + 56 = 0 \] has radius 3.

• A. \( 7, -7 \)
• B. \( 9, -9 \)
• C. \( 12, -12 \)
• D. \( 18, -18 \)
• E. \( 14, -14 \)

Hint:

Use the standard form of the circle equation and apply the radius formula to solve.

Answer 1

Answer: (E)

The equation of a circle: \[ x^2 + y^2 + Dx + Ey + F = 0 \] has center \( (-D/2, -E/2) \) and radius: \[ r = \sqrt{\left( \frac{D}{2} \right)^2 + \left( \frac{E}{2} \right)^2 - F} \] Comparing: \[ D = \alpha, \quad E = -8, \quad F = 56 \] The radius is 3: \[ \sqrt{\left( \frac{\alpha}{2} \right)^2 + \left( \frac{-8}{2} \right)^2 - 56} = 3 \] \[ \left( \frac{\alpha}{2} \right)^2 + (-4)^2 - 56 = 9 \] \[ \frac{\alpha^2}{4} + 16 - 56 = 9 \] \[ \frac{\alpha^2}{4} - 40 = 9 \] \[ \frac{\alpha^2}{4} = 49 \] \[ \alpha^2 = 196 \] \[ \alpha = \pm 14 \] Final Answer: \[ \boxed{14, -14} \]