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 4√2 ,…
JEE Main 2024 — Mathematics Coordinate Geometry
2024integermedium
Consider a circle (x−α2)+(y−β2)=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 _______.
Official previous-year question
Held on 27 Jan 2024 · Verified 6 Jul 2026.
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Solution
Given: (x−α2)+(y−β2)=50
The given equation of circle represent centre as (α,β) and radius as 52 units.
Now, x+y=0 is tangent to the given circle at P.
We know that, radius is perpendicular to tangent at the point of tangency.
⇒CP⊥(x+y=0) and CP=r
⇒r=∣12+12α×1+β×1∣
⇒52=∣2α+β∣
⇒50=2(α+β)2
⇒(α+β)2=100
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