If the circles x^2 + y^2 = 16 and x^2 + y^2 + 2ax + 4y + 4 = 0 touch each other internally then a =

Step-by-step solution

For the first circle x^2 + y^2 = 16 , the center is C₁(0, 0) and the radius is r₁ = 4 . For the second circle x^2 + y^2 + 2ax + 4y + 4 = 0 , the center is C₂(-a, -2) and the radius is r₂ = a^2 + 2^2 - 4 = |a| . Since the two circles touch each other internally, the distance between their centers is equal to the difference of their radii: C₁C₂ = |r₁ - r₂| (-a - 0)^2 + (-2 - 0)^2 = |4 - |a|| a^2 + 4 = |4 - |a|| Squaring both sides: a^2 + 4 = 16 + a^2 - 8|a| 8|a| = 12 |a| = 3 2 a = 3 2 From the given options, a = 3 2