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

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Matrices

Let

A = 143 2 - 2 8 78 - 7

and

det (A - α I) = 0

, where

α

is a real number. If the largest possible value of

α

p

, then the circle

(x - p)^{2} + (y - 2 p)^{2} = 320

, intersects the co-ordinate axes at

Answer: (C)

The characteristic equation

det (A - α I) = 0

expands to

α^{3} + 8 α^{2} - 88 α - 320 = 0

, which factors as

(α - 8) (α^{2} + 16 α + 40) = 0

; the quadratic has roots

- 8 \pm 26

, so the largest eigenvalue is

p = 8

. The circle is then

(x - 8)^{2} + (y - 16)^{2} = 320

. Since

8^{2} + 1 6^{2} = 320

, it passes through the origin. Setting

y = 0

(x - 8)^{2} = 64

gives

x = 0, 16

; setting

x = 0

(y - 16)^{2} = 256

gives

y = 0, 32

. The distinct points on the axes are

(0, 0), (16, 0), (0, 32)

, so the circle meets the coordinate axes at

3

points.

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Single correct option