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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(x-{\alpha }^{2})+(y-{\beta }^{2})=50(x−α2)+(y−β2)=50, where α,β>0\alpha ,\beta >0α,β>0. If the circle touches the line y+x=0y+x=0y+x=0 at the point PPP, whose distance from the origin is 424\sqrt{2}42​ , then (α+β)2(\alpha +\beta {)}^{2}(α+β)2 is equal to _______.

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Held on 27 Jan 2024 · Verified 6 Jul 2026.

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