JEE Main 2024MathematicsCircleMediumNumerical

JEE Main 2024 — Circle Question with Solution

JEE Main 2024 (27 Jan Shift 2)

Question

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 _______.

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Show full solutionCorrect answer: 100
Correct answer
100

Step-by-step explanation

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=α×1+β×112+12

⇒52=α+β2

⇒50=α+β22

⇒α+β2=100

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About this question

This is a previous-year question from JEE Main 2024, covering the Circle chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.