The circle x2+y2+6x+8y+16=0 has centre (−3,−4) and radius 3 units, The circle x2+y2+2(3−√3)x+2(4−√6)y=k+6√3+8√6,k>0 has centre (√3−3,√6−4) and radius √k+34 ∵ These two circles touch internally hence √3+6=|√k+34−3| Here, k=2 is only possible (∵k>0) Equation of common tangent to two circles is 2√3x+2√6y+16+6√3+8√6+k=0 ∵k=2 then equation is x+√2y+3+4√2+3√3=0...... (i) ∵(α,β) are foot of perpendicular from (−3,−4) To line (i) then
α+3
1
=
β+4
√2
=
−(−3−4√2+3+4√2+3√3)
1+2
∴α+3=
β+4
√2
=−√3 ⇒(α+√3)2=9 and (β+√6)2=16 ∴(α+√3)2+(β+√6)2=25