

∵316+α=6 and 315+β=6⇒(α,β)≡(2,3) Also, C1C2=r1+r2 ⇒(2−8)2+(3−215)2=2r2+r2⇒r2=25⇒r1=2r2=5∴(α+β)+4(r12+r22)=5+4(425+25)=130
Let the circles C1:(x−α)2+(y−β)2=r12 and C2:(x−8)2+(y−215)2=r22 touch each other externally at the point (6,6). If the point (6,6) divides the line segment joining the centres of the circles C1 and C2 internally in the ratio 2:1, then (α+β)+4(r12+r22) equals
Held on 8 Apr 2024 · Verified 6 Jul 2026.
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