Consider a circle (x−α)2+(y−β)2=50, where α,β>0. If the circle touches the line y+x=0 at the point P, whose distance from the origin is 42, then (α+β)2 is equal to
Answer: 100
Solution
The circle (x−α)2+(y−β)2=50 has centre (α,β) and radius r=50=52.
The circle touches the line x+y=0, so the perpendicular distance from the centre to this line equals the radius:
2∣α+β∣=52⇒∣α+β∣=10.
Since α,β>0, α+β=10 (the point of tangency at distance 42 from the origin fixes α and β individually, but this sum is all that's needed here).
(α+β)2=100.
JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Integer answer
