JEE Main 2026 · 6 Apr · Shift 2 · Mathematics Q23

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MathematicsInteger+4 / -1Question 23 of 75

Let the line $x-y=4$ intersect the circle $C:(x-4)^2+(y+3)^2=9$ at the points Q and R. If $P(\alpha,\beta)$ is a point on C such that $PQ=PR$ , then $(6\alpha+8\beta)^2$ is equal to ________.

Your answer

Answer and worked solution

Official answer: 18

P on the perpendicular bisector of QR. That line passes through the centre $(4,-3)$ and is perpendicular to $x-y=4$ , so it has direction $(1,-1)$ .

Points. $P=(4,-3)\pm\dfrac3{\sqrt2}(1,-1)$ .

Value. $6\alpha+8\beta=24-24\pm\dfrac3{\sqrt2}(6-8)=\mp3\sqrt2$ , whose square is 18.

Answer: 18

Topics:Equation of a circle Equations of a line

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JEE Main 2026 · 6 Apr · Shift 2, Mathematics question 23