If the circles x 2 + y 2 + 6 x + 8 y + 16 = 0 and x 2 + y 2 + 2 3 - 3 x + 2 4 - 6 y = k + 6 3 + 8 6 , k > 0 , touch internally at the point P α , β , then α + 3 2 + β + 6 2 is equal to _______.

Step-by-step solution

The circle x 2 + y 2 + 6 x + 8 y + 16 = 0 has centre - 3 , - 4 and radius 9 + 16 - 16 = 3 units. The circle x 2 + y 2 + 2 3 - 3 x + 2 4 - 6 y = k + 6 3 + 8 6 , k > 0 has centre 3 - 3 , 6 - 4 and radius 3 - 3 2 + 6 - 4 2 + k + 6 3 + 8 6 = k + 34 Given that these two circles touch internally, so distance between their centres = difference of radii 3 + 6 = k + 34 - 3 ⇒ k + 34 - 3 = ± 3 Here, k = 2 is only possible value ∵     k > 0 Now the equation of common tangent to both