JEE Main 2026 — Mathematics Algebra

The characteristic equation is given by $\det(A-\alpha I) = 0$det(A−αI)=0.

$$\det \begin{bmatrix} 1-\alpha & 2 & 7 \\ 4 & -2-\alpha & 8 \\ 3 & 8 & -7-\alpha \end{bmatrix} = 0$$det​1−α43​2−2−α8​78−7−α​​=0

Expanding the determinant along the first row:

$$(1-\alpha)[(-2-\alpha)(-7-\alpha) - 64] - 2[4(-7-\alpha) - 24] + 7[32 - 3(-2-\alpha)] = 0$$(1−α)[(−2−α)(−7−α)−64]−2[4(−7−α)−24]+7[32−3(−2−α)]=0

$$(1-\alpha)(\alpha^2 + 9\alpha - 50) - 2(-4\alpha - 52) + 7(3\alpha + 38) = 0$$(1−α)(α2+9α−50)−2(−4α−52)+7(3α+38)=0

$$-\alpha^3 - 8\alpha^2 + 59\alpha - 50 + 8\alpha + 104 + 21\alpha + 266 = 0$$−α3−8α2+59α−50+8α+104+21α+266=0

$$\alpha^3 + 8\alpha^2 - 88\alpha - 320 = 0$$α3+8α2−88α−320=0

By inspection, $\alpha = 8$α=8 is a root since $8^3 + 8(8^2) - 88(8) - 320 = 512 + 512 - 704 - 320 = 0$83+8(82)−88(8)−320=512+512−704−320=0.

Factoring out $(\alpha - 8)$(α−8), we get:

$$(\alpha - 8)(\alpha^2 + 16\alpha + 40) = 0$$(α−8)(α2+16α+40)=0

The roots are $\alpha = 8$α=8 and $\alpha = \dfrac{-16 \pm \sqrt{256 - 160}}{2} = -8 \pm 2\sqrt{6}$α=2−16±256−160​​=−8±26​.

The largest possible value of $\alpha$α is $p = 8$p=8.

Substituting $p = 8$p=8 into the given equation of the circle:

$$(x-8)^2 + (y-16)^2 = 320$$(x−8)2+(y−16)2=320

To find the intersection points with the x-axis, substitute $y = 0$y=0:

$$(x-8)^2 + (-16)^2 = 320$$(x−8)2+(−16)2=320

$$(x-8)^2 + 256 = 320 \Rightarrow (x-8)^2 = 64 \Rightarrow x-8 = \pm 8$$(x−8)2+256=320⇒(x−8)2=64⇒x−8=±8

$x = 0$x=0 or $x = 16$x=16

The circle intersects the x-axis at $(0, 0)$(0,0) and $(16, 0)$(16,0).

To find the intersection points with the y-axis, substitute $x = 0$x=0:

$$(-8)^2 + (y-16)^2 = 320$$(−8)2+(y−16)2=320

$$64 + (y-16)^2 = 320 \Rightarrow (y-16)^2 = 256 \Rightarrow y-16 = \pm 16$$64+(y−16)2=320⇒(y−16)2=256⇒y−16=±16

$y = 0$y=0 or $y = 32$y=32

The circle intersects the y-axis at $(0, 0)$(0,0) and $(0, 32)$(0,32).

The distinct points of intersection with the coordinate axes are $(0, 0)$(0,0), $(16, 0)$(16,0), and $(0, 32)$(0,32).

Thus, the circle intersects the coordinate axes at exactly $3$3 points.

Answer: $3$3 points