JEE Main 2026 · 4 Apr · Shift 2 · Mathematics Q6

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Let $A=\begin{bmatrix}1&2&7\\4&-2&8\\3&8&-7\end{bmatrix}$ and $\det(A-\alpha I)=0$ , where $\alpha$ is a real number. If the largest possible value of $\alpha$ is $p$ , then the circle $(x-p)^2+(y-2p)^2=320$ , intersects the co-ordinate axes at

1 point

2 points

3 points

4 points

Answer and worked solution

Official answer: C

Eigenvalues. The characteristic polynomial is $\alpha^3+8\alpha^2-88\alpha-320=(\alpha-8)(\alpha^2+16\alpha+40)$ . The largest root is $p=8$ .

Circle. $(x-8)^2+(y-16)^2=320$ has radius $\sqrt{320}\approx17.9$ .

It passes through the origin: $64+256=320$ .
Each axis is cut twice, since the distances from the centre (16 and 8) are less than the radius.
The origin is common to both axes.

3 points.

(C)

Topics:Determinants and properties Equation of a circle

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JEE Main 2026 · 4 Apr · Shift 2, Mathematics question 6