Question
Mathematics Question on circle
If the circles x2+y2+6x+8y+16=0 and x2+y2+2(3−3)x+x+2(4−6)y = k +6 \sqrt{3}+8 \sqrt{6}, k >0; touch internally at the point P(α,β), then (α+3)2+(β+6)2 is equal to _______
Answer
The correct answer is 25
C1(−3,−4)
r1=25−16=3
C2=(−3+3,−4+6)
r2=34+k
C1C2=∣r1−r2∣
C1C2=3+6=3
3=∣3−34+k∣⇒k=2
r2=6
(α,β)=(−3−3,−4−6)
(α+3)2+(β+6)2=9+16=25